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Properties of definite integrals

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54909912
The function \(f\) is defined by one formula on \([-2, 1]\) and another on \((1, 5]\). Write \(\int_{-2}^{5}f(x)\,\text{d}x\) as a sum of two definite integrals that respects the change in formula.

Hints

- Locate the point where the defining rule changes. - Use that point as the shared endpoint of two adjacent intervals. - Preserve the original direction from the lower bound to the upper bound.

Solution

1. The formula changes at \(x=1\), so split the interval there. 2. Add the accumulated values over the adjacent intervals. 3. The result is \(\int_{-2}^{1}f(x)\,\text{d}x+\int_{1}^{5}f(x)\,\text{d}x\).

Answer

\(\int_{-2}^{1}f(x)\,\text{d}x+\int_{1}^{5}f(x)\,\text{d}x\)
55598512
Suppose \(\int_2^7 f(x)\,\text{d}x=9\). Find \(\int_7^2 f(x)\,\text{d}x\) and \(\int_2^2 f(x)\,\text{d}x\) without finding a formula for \(f\).

Hints

- Compare the order of the limits in the first requested integral with the given one. - Think about the signed interval length when both limits are the same.

Solution

1. Reversing the limits changes the sign, so \(\int_7^2 f(x)\,\text{d}x=-9\). 2. An integral with identical upper and lower limits is zero, so \(\int_2^2 f(x)\,\text{d}x=0\).

Answer

\(\int_7^2 f(x)\,\text{d}x=-9\) and \(\int_2^2 f(x)\,\text{d}x=0\).
52456112
Justify each equation without evaluating the integrals directly. Use properties of definite integrals or symmetry. a) \(\int_{-5}^{5} (x^3 + x) \, \text{d}x = 0\) b) \(\int_{1}^{4} f(x) \, \text{d}x + \int_{4}^{1} f(x) \, \text{d}x = 0\) c) \(\int_{a}^{a} \sqrt{x^2 + 1} \, \text{d}x = 0\)

Hints

- Determine how the graph of an odd function behaves under a rotation of \(180^\circ\) about the origin. - What happens to the sign of a definite integral when its limits are reversed? - What is the width of the interval when the lower and upper limits are equal? - Recall that a definite integral represents net signed area.

Solution

1. For a), let \(g(x)=x^3+x\). Since \(g(-x)=(-x)^3+(-x)=-x^3-x=-g(x)\), the integrand is odd. The signed areas on the symmetric interval \([-5, 5]\) cancel, so the integral is \(0\). 2. For b), reversing the limits changes the sign of a definite integral: \(\int_a^b f(x)\,\text{d}x=-\int_b^a f(x)\,\text{d}x\). Therefore, \(\int_1^4 f(x)\,\text{d}x+\int_4^1 f(x)\,\text{d}x=\int_1^4 f(x)\,\text{d}x-\int_1^4 f(x)\,\text{d}x=0\). 3. For c), a definite integral whose lower and upper limits are equal is always \(0\): \(\int_a^a h(x)\,\text{d}x=0\).

Answer

a) The integrand is odd and the limits are symmetric about \(0\), so the signed areas cancel. b) Reversing the limits changes the sign of the integral, so the two integrals cancel. c) A definite integral with identical limits is \(0\).
52459412
A continuous function \(f\) has period \(2\pi\), and \(\int_0^{2\pi} f(x)\,\text{d}x=6\). Without finding an antiderivative, determine \(\int_a^{a+4\pi} f(x)\,\text{d}x\) for an arbitrary real number \(a\). Explain which properties of definite integrals make the value independent of \(a\).

Hints

- Break the interval of length \(4\pi\) into two adjacent intervals of length \(2\pi\). - A periodic function has the same accumulated signed area over every full period. - Combine the two period integrals using additivity.

Solution

1. Because \(f\) has period \(2\pi\), the integral over any interval of length \(2\pi\) equals the integral over \([0,2\pi]\), so \(\int_a^{a+2\pi}f(x)\,\text{d}x=6\). 2. The next interval \([a+2\pi,a+4\pi]\) is another full period, so its integral is also \(6\). 3. By additivity over adjacent intervals, \(\int_a^{a+4\pi}f(x)\,\text{d}x=6+6=12\). 4. Thus the value is independent of \(a\) because a horizontal shift by a whole period preserves the integral over one period.

Answer

For every real \(a\), \(\int_a^{a+4\pi}f(x)\,\text{d}x=12\).
52460212
An integrable function \(g\) satisfies \(\int_{-1}^{2}g(x)\,\text{d}x=5\) and \(\int_{2}^{6}g(x)\,\text{d}x=-5\). Without finding a formula for \(g\) or an antiderivative, find a nontrivial interval with endpoints chosen from \(-1\), \(2\), and \(6\) on which the integral of \(g\) is \(0\). Then give the value of the integral over the same interval with the limits reversed.

Hints

- Look for adjacent intervals whose signed accumulations cancel. - Use additivity before doing anything with reversed limits. - Reversing the limits negates the integral.

Solution

1. By additivity, \(\int_{-1}^{6}g(x)\,\text{d}x=\int_{-1}^{2}g(x)\,\text{d}x+\int_{2}^{6}g(x)\,\text{d}x\). 2. Therefore \(\int_{-1}^{6}g(x)\,\text{d}x=5+(-5)=0\), so \([-1,6]\) is the required nontrivial interval. 3. Reversing the limits changes the sign, and the negative of \(0\) is still \(0\). Hence \(\int_6^{-1}g(x)\,\text{d}x=0\).

Answer

The interval is \([-1,6]\), with \(\int_{-1}^{6}g(x)\,\text{d}x=0\). Also, \(\int_6^{-1}g(x)\,\text{d}x=0\).
52461312
Justify each statement without evaluating the integrals. Use even or odd symmetry of the integrand. a) \(\int_{-a}^{a} (x^3-x) \, \text{d}x=0\) for every \(a>0\) b) \(\int_{-2}^{2} \cos(x) \, \text{d}x=2\int_{0}^{2} \cos(x) \, \text{d}x\) c) \(\int_{-1}^{1} \sin(x)\cos(x) \, \text{d}x=0\)

Hints

- Test each integrand by replacing \(x\) with \(-x\). - What is the integral of an odd function over a symmetric interval? - How are the two halves of a symmetric interval related for an even function? - Recall how the parity of a product depends on the parity of its factors.

Solution

1. For a), let \(f(x)=x^3-x\). Since \(f(-x)=(-x)^3-(-x)=-x^3+x=-f(x)\), the integrand is odd. The signed areas on the symmetric interval \([-a, a]\) cancel, so the integral is \(0\). 2. For b), \(g(x)=\cos(x)\) is even because \(g(-x)=\cos(-x)=\cos(x)=g(x)\). Therefore, the signed areas on \([-2, 0]\) and \([0, 2]\) are equal, giving twice the integral on \([0, 2]\). 3. For c), sine is odd and cosine is even, so their product is odd: \(h(-x)=\sin(-x)\cos(-x)=-\sin(x)\cos(x)=-h(x)\). The integral of an odd function over \([-1, 1]\) is \(0\).

Answer

a) The integrand is odd, so its signed areas cancel over \([-a, a]\). b) The integrand is even, so the two halves of the integral are equal. c) The product of an odd function and an even function is odd, so the integral over \([-1, 1]\) is \(0\).
52462412
The graph of \(f\) is shown. Use its graphical symmetry to justify the equation \(\int_1^2 f(x)\,\text{d}x=\int_2^3 f(x)\,\text{d}x\).
Figure for problem 524624

Hints

- Identify the vertex and axis of symmetry from the graph. - Compare the positions and widths of the two intervals. - Ask how mirrored graph heights affect the two accumulated areas.

Solution

1. The displayed parabola has vertex \((2,0)\) and axis of symmetry \(x=2\). 2. The intervals \([1,2]\) and \([2,3]\) have the same width and are mirror images across \(x=2\). 3. Corresponding function values are equal on those mirrored intervals, so the areas under the graph are equal. Therefore the two definite integrals are equal.

Answer

The graph is symmetric about \(x=2\), and \([1,2]\) and \([2,3]\) are mirror-image intervals across that line. Therefore \(\int_1^2 f(x)\,\text{d}x=\int_2^3 f(x)\,\text{d}x\).
52508112
Justify each equation using symmetry, linearity, or reversed limits without fully evaluating the nonconstant integrals. a) \(\int_{0}^{2}(x-1)^5 \, \text{d}x=0\) b) \(\int_{1}^{e}\frac{1}{x} \, \text{d}x+\int_{e}^{1}\frac{1}{x} \, \text{d}x=0\) c) \(\int_{0}^{2}(x^2+4) \, \text{d}x=\int_{0}^{2}x^2 \, \text{d}x+8\)

Hints

- Compare the first graph with the midpoint of its interval. - What happens to an integral when its limits are reversed? - Split the integral of a sum into a sum of integrals. - Interpret the integral of a constant geometrically.

Solution

1. For a), the graph of \(y=(x-1)^5\) has point symmetry about \((1, 0)\), and the interval \([0, 2]\) is symmetric about \(x=1\). The signed areas cancel, so the integral is \(0\). 2. For b), reversing the limits changes the sign: \(\int_e^1\frac{1}{x}\,\text{d}x=-\int_1^e\frac{1}{x}\,\text{d}x\). Their sum is \(0\). 3. For c), linearity gives \(\int_0^2(x^2+4)\,\text{d}x=\int_0^2x^2\,\text{d}x+\int_0^2 4\,\text{d}x\). The constant integral is the area of a rectangle with width \(2\) and height \(4\), so it equals \(8\).

Answer

a) Point symmetry about \((1, 0)\) makes the signed areas on the two halves cancel. b) The second integral is the negative of the first because its limits are reversed. c) By linearity, the constant term contributes \(\int_0^2 4\,\text{d}x=8\).
52956312
Let \(f\) and \(g\) be continuous on \(\mathbb{R}\). Suppose \(\int_{2}^{6}f(x) \, \text{d}x=12\) and \(\int_{2}^{6}g(x) \, \text{d}x=5\). Use linearity to evaluate \(\int_{2}^{6}[1.5f(x)-4g(x)+3] \, \text{d}x\).

Hints

- Split the integral into separate terms. - Move constant factors outside their integrals. - Integrate the constant term over the full interval. - Substitute the two given integral values.

Solution

1. Apply linearity: \(\int_2^6[1.5f(x)-4g(x)+3]\,\text{d}x=1.5\int_2^6f(x)\,\text{d}x-4\int_2^6g(x)\,\text{d}x+\int_2^6 3\,\text{d}x\). 2. Substitute the known values and evaluate the constant integral: \(1.5\cdot 12-4\cdot 5+3(6-2)=18-20+12\). 3. The result is \(10\).

Answer

\(10\)
52956412
Let \(h\) be continuous on \([1, 3]\), and suppose \(\int_1^3h(x)\,\text{d}x=A\). Find the real constant \(c\), in terms of \(A\), such that \(\int_1^3(2h(x)-c)\,\text{d}x=0\).

Hints

- Use the sum and constant-multiple properties. - The integral of a constant over an interval equals the constant times the interval length. - Substitute the known integral value. - Solve the resulting linear equation.

Solution

1. Use linearity: \(\int_1^3(2h(x)-c)\,\text{d}x=2\int_1^3h(x)\,\text{d}x-\int_1^3c\,\text{d}x\). 2. Substitute the known value: \(2A-c(3-1)=0\). 3. Thus, \(2A-2c=0\), so \(c=A\).

Answer

\(c=A\)
52956512
For \(k>1\), let \(A=\int_1^k x^3\,\text{d}x\) and \(B=\int_1^k\frac{1}{2}x^3\,\text{d}x\). Without evaluating either integral, use a property of definite integrals to express \(B\) in terms of \(A\) and find \(\frac{B}{A}\). Explain why the ratio is defined.

Hints

- Compare the two integrands before trying to integrate them. - A constant factor can be moved outside a definite integral. - Use the sign of \(x^3\) on \([1,k]\) to decide whether \(A\) can be zero.

Solution

1. By the constant-multiple property, \(B=\frac{1}{2}\int_1^k x^3\,\text{d}x=\frac{1}{2}A\). 2. Because \(k>1\) and \(x^3>0\) on \([1,k]\), \(A>0\), so the ratio is defined. 3. Therefore \(\frac{B}{A}=\frac{1}{2}\).

Answer

\(B=\frac{1}{2}A\) and \(\frac{B}{A}=\frac{1}{2}\). The ratio is defined because \(A>0\).
52957512
Suppose \(\int_1^5 f(x) \, \text{d}x=8\) and \(\int_1^5 g(x) \, \text{d}x=-2\). Evaluate \(I=\int_5^1[3f(x)+4g(x)] \, \text{d}x\).

Hints

- Reverse the limits so they match the given integrals. - Use linearity to separate the terms. - Substitute the two known integral values.

Solution

1. Reverse the limits: \(I=-\int_1^5[3f(x)+4g(x)]\,\text{d}x\). 2. Apply linearity: \(I=-\left(3\int_1^5f(x)\,\text{d}x+4\int_1^5g(x)\,\text{d}x\right)\). 3. Substitute the known values: \(I=-[3\cdot 8+4\cdot(-2)]=-16\).

Answer

\(I=-16\)
52957812
Combine the difference into one integral, then evaluate it: \(\int_{-1}^{3}(4-x^2) \, \text{d}x-\int_{2}^{3}(4-x^2) \, \text{d}x\).

Hints

- Interpret the subtraction as removing a subinterval. - Reverse the limits of the subtracted integral if helpful. - Identify the remaining interval before finding an antiderivative. - Evaluate the polynomial antiderivative at both endpoints.

Solution

1. Subtracting the second integral removes the contribution over \([2, 3]\): \(\int_{-1}^{3}(4-x^2)\,\text{d}x-\int_{2}^{3}(4-x^2)\,\text{d}x=\int_{-1}^{2}(4-x^2)\,\text{d}x\). 2. An antiderivative is \(F(x)=4x-\frac{1}{3}x^3\). 3. Evaluate: \(F(2)-F(-1)=\frac{16}{3}-(-\frac{11}{3})=9\).

Answer

\(9\)
52957912
Use additivity over intervals to combine the expression into one integral, then evaluate it: \(\int_{-1}^{2}(3x^2-4x+5) \, \text{d}x-\int_{4}^{2}(3x^2-4x+5) \, \text{d}x+\int_{4}^{5}(3x^2-4x+5) \, \text{d}x\).

Hints

- Eliminate the minus sign by reversing the middle limits. - Check whether the resulting intervals form one continuous chain. - Simplify the integral expression before finding an antiderivative.

Solution

1. Reverse the limits in the middle term: \(-\int_4^2f(x)\,\text{d}x=\int_2^4f(x)\,\text{d}x\), where \(f(x)=3x^2-4x+5\). 2. The intervals \([-1, 2]\), \([2, 4]\), and \([4, 5]\) join to form \([-1, 5]\). Thus the expression is \(\int_{-1}^{5}(3x^2-4x+5)\,\text{d}x\). 3. An antiderivative is \(F(x)=x^3-2x^2+5x\). 4. Evaluate: \(F(5)-F(-1)=100-(-8)=108\).

Answer

\(\int_{-1}^{5}(3x^2-4x+5) \, \text{d}x=108\)
52958012
Use linearity to simplify the expression as much as possible, then evaluate it: \(3\int_{1}^{4}\left(\sqrt{x}+\frac{1}{x^2}\right) \, \text{d}x+\int_{1}^{4}(2-3\sqrt{x}) \, \text{d}x\).

Hints

- Combine the two integrals because they have the same limits. - Distribute the constant factor before adding the integrands. - Identify terms that cancel. - Rewrite reciprocal powers using negative exponents when integrating.

Solution

1. Combine the integrals over the same interval: \(\int_1^4\left[3\left(\sqrt{x}+\frac{1}{x^2}\right)+2-3\sqrt{x}\right]\,\text{d}x\). 2. Simplify the integrand: \(3\sqrt{x}+\frac{3}{x^2}+2-3\sqrt{x}=\frac{3}{x^2}+2\). 3. An antiderivative is \(F(x)=-\frac{3}{x}+2x\). 4. Evaluate: \(F(4)-F(1)=\frac{29}{4}-(-1)=\frac{33}{4}\).

Answer

\(\int_{1}^{4}\left(\frac{3}{x^2}+2\right) \, \text{d}x=\frac{33}{4}\)
52959312
For \(k>0\), evaluate each integral using symmetry to minimize computation. a) \(\int_{-k}^{k}(10x^9-4x^5+x) \, \text{d}x\) b) \(\int_{-k}^{k}\left(\frac{1}{4}x^4-2\right) \, \text{d}x\)

Hints

- Determine whether each integrand is even or odd. - What is the integral of an odd function over \([-k, k]\)? - For an even function, integrate over \([0, k]\) and double the result.

Solution

1. For a), the integrand is odd, so its integral over \([-k, k]\) is \(0\). 2. For b), the integrand is even. Therefore, \(\int_{-k}^{k}(\frac{1}{4}x^4-2)\,\text{d}x=2\int_0^k(\frac{1}{4}x^4-2)\,\text{d}x\). 3. An antiderivative is \(\frac{1}{20}x^5-2x\), so the value is \(2[\frac{1}{20}x^5-2x]_0^k=\frac{1}{10}k^5-4k\).

Answer

a) \(0\) b) \(\frac{1}{10}k^5-4k\)
52961212
Let \(h\) be an integrable odd function on \([-5,5]\), and define \(f(x)=\frac{3}{10}x^4+h(x)\). Evaluate \(\int_{-5}^{5}f(x)\,\text{d}x\). Your reasoning must use the symmetry of \(h\); no formula for \(h\) is available.

Hints

- Use the given parity of \(h\) before considering the polynomial term. - An odd function contributes zero net integral on a symmetric interval. - The polynomial term is even, so symmetry can also shorten that calculation.

Solution

1. Because \(h\) is odd, \(\int_{-5}^{5}h(x)\,\text{d}x=0\). 2. The remaining term is even, so \(\int_{-5}^{5}\frac{3}{10}x^4\,\text{d}x=2\int_0^5\frac{3}{10}x^4\,\text{d}x\). 3. An antiderivative of \(\frac{3}{10}x^4\) is \(\frac{3}{50}x^5\). 4. Therefore \(2\left[\frac{3}{50}x^5\right]_0^5=375\).

Answer

\(375\)
52969312
Evaluate each definite integral, using symmetry when possible. a) \(\int_{-5}^{5}(x^3-25x) \, \text{d}x\) b) \(\int_{-2}^{2}(3x^2-12) \, \text{d}x\)

Hints

- Inspect the powers of \(x\) in each integrand. - Recall the integral rules for even and odd functions over \([-a, a]\). - Symmetry may eliminate or halve the computation. - Decide whether each graph is symmetric about the origin or the y-axis.

Solution

1. For a), the integrand is odd, so its integral over \([-5, 5]\) is \(0\). 2. For b), the integrand is even. Therefore, \(\int_{-2}^{2}(3x^2-12)\,\text{d}x=2\int_0^2(3x^2-12)\,\text{d}x\). 3. An antiderivative is \(x^3-12x\), so the value is \(2[x^3-12x]_0^2=2(8-24)=-32\).

Answer

a) \(0\) b) \(-32\)
52979512
Evaluate \(\int_1^2 5x^4\,\text{d}x\). Show how the constant multiple rule lets you move the factor \(5\) outside the integral.

Hints

- Identify the constant factor in the integrand. - Move that factor outside the integral. - Use an antiderivative of \(x^4\), then evaluate at the bounds.

Solution

1. Apply the constant multiple rule: \(\int_1^2 5x^4\,\text{d}x=5\int_1^2x^4\,\text{d}x\). 2. An antiderivative of \(x^4\) is \(\frac{1}{5}x^5\), so \(5\left[\frac{1}{5}x^5\right]_1^2=[x^5]_1^2\). 3. Evaluate the bounds: \(2^5-1^5=32-1=31\).

Answer

\(31\)
52979612
A continuous function \(g\) satisfies \(\int_2^6 g(x)\,\text{d}x=18\). Use the constant multiple rule to evaluate \(\int_2^6\frac{2}{3}g(x)\,\text{d}x\). Briefly justify your step.

Hints

- You do not need a formula for \(g\). - A constant factor can be moved outside a definite integral. - Multiply the given integral value by \(\frac{2}{3}\).

Solution

1. By the constant multiple rule, \(\int_2^6\frac{2}{3}g(x)\,\text{d}x=\frac{2}{3}\int_2^6g(x)\,\text{d}x\). 2. Substitute the given value: \(\frac{2}{3}\cdot 18=12\).

Answer

\(12\), because \(\int_2^6\frac{2}{3}g(x)\,\text{d}x=\frac{2}{3}\int_2^6g(x)\,\text{d}x\).
52979912
Evaluate \(\int_1^2\left(6x^2-\frac{4}{x^2}\right)\,\text{d}x\). Use the sum and constant multiple rules to build an antiderivative from the two terms.

Hints

- Split the integral into two simpler integrals. - Rewrite \(\frac{1}{x^2}\) using a negative exponent. - After finding an antiderivative, evaluate it at the upper and lower bounds.

Solution

1. Split the integral and rewrite the reciprocal power: \(\int_1^2 6x^2\,\text{d}x-4\int_1^2x^{-2}\,\text{d}x\). 2. An antiderivative of the integrand is \(F(x)=2x^3+\frac{4}{x}\). 3. Apply the Fundamental Theorem of Calculus: \(F(2)-F(1)=\left(16+2\right)-\left(2+4\right)=18-6=12\).

Answer

\(12\)
53268912
The graph shown represents a piecewise linear function \(f\). The three regions between the graph and the x-axis have areas \(A_1=4\), \(A_2=6\), and \(A_3=2\). Evaluate each definite integral. a) \(\int_{-4}^{-1}f(x)\,\text{d}x\) b) \(\int_{-1}^{3}f(x)\,\text{d}x\) c) \(\int_{-4}^{3}f(x)\,\text{d}x\) d) \(\int_{-4}^{5}f(x)\,\text{d}x\)
Figure for problem 532689

Hints

- Determine whether each labeled region lies above or below the x-axis. - Use positive contributions above the axis and negative contributions below it. - Apply interval additivity when an integral spans more than one labeled region.

Solution

1. A region above the x-axis contributes its area positively to a definite integral. A region below the axis contributes the negative of its area. 2. For a), \(A_1\) lies below the axis, so \(\int_{-4}^{-1}f(x)\,\text{d}x=-A_1=-4\). 3. For b), \(A_2\) lies above the axis, so \(\int_{-1}^{3}f(x)\,\text{d}x=A_2=6\). 4. For c), use interval additivity: \(\int_{-4}^{3}f(x)\,\text{d}x=-A_1+A_2=-4+6=2\). 5. For d), include all three regions: \(\int_{-4}^{5}f(x)\,\text{d}x=-A_1+A_2-A_3=-4+6-2=0\).

Answer

a) \(-4\) b) \(6\) c) \(2\) d) \(0\)
53465712
The graph of \(f\) is shown on \([-3,3]\). Without finding an antiderivative, explain why \(\int_{-3}^{3}f(x)\,\text{d}x=0\). Refer explicitly to the graph's symmetry and net signed area.
Figure for problem 534657

Hints

- Inspect what happens to the graph under a \(180^\circ\) rotation about the origin. - Compare corresponding areas to the left and right of the y-axis. - Recall how a region below the x-axis contributes to a definite integral.

Solution

1. The displayed graph has point symmetry about the origin, so the areas on corresponding intervals to the left and right of the y-axis have equal magnitude. 2. The graph lies above the x-axis on one side exactly where its origin-symmetric counterpart lies below the x-axis on the other side. 3. Therefore the equal-magnitude areas contribute with opposite signs and cancel, giving \(\int_{-3}^{3}f(x)\,\text{d}x=0\).

Answer

The displayed graph has point symmetry about the origin, so equal-magnitude areas on the two sides of the y-axis have opposite signs. Their signed contributions cancel, and \(\int_{-3}^{3}f(x)\,\text{d}x=0\).
54910112
The net signed accumulation of \(f\) from \(x=3\) to \(x=9\) is \(42\). Write three equivalent definite-integral equations: a) one integral over the full interval, b) a sum of two integrals split at \(x=6\), and c) one integral with the limits reversed.

Hints

- Use the stated starting and ending x-values as the direct bounds. - Split an integral by using the same interior point as the upper bound of one part and the lower bound of the next. - Reversing the order of integration changes the sign. - Check that all three equations describe the same oriented accumulation.

Solution

1. a) The direct equation is \(\int_3^9f(x)\,\text{d}x=42\). 2. b) Additivity over adjacent intervals gives \(\int_3^6f(x)\,\text{d}x+\int_6^9f(x)\,\text{d}x=42\). 3. c) Reversing the limits changes the sign, so \(\int_9^3f(x)\,\text{d}x=-42\).

Answer

a) \(\int_3^9f(x)\,\text{d}x=42\) b) \(\int_3^6f(x)\,\text{d}x+\int_6^9f(x)\,\text{d}x=42\) c) \(\int_9^3f(x)\,\text{d}x=-42\)
54910512
A room's temperature changes at rate \(r(t)\) degrees Fahrenheit per hour, where \(t=0\) is noon and \(t=3\) is \(3{:}00\) p.m. Write two equivalent definite-integral expressions for “the temperature at noon minus the temperature at \(3{:}00\) p.m.”

Hints

- First express the forward-time temperature change. - Reverse the order of the temperature subtraction by changing a sign. - Use reversed integration limits as another way to encode that sign change.

Solution

1. The forward change is \(T(3)-T(0)=\int_0^3r(t)\,\text{d}t\). 2. Reversing the subtraction gives \(T(0)-T(3)=-\int_0^3r(t)\,\text{d}t\). 3. Reversing the limits gives the equivalent expression \(\int_3^0r(t)\,\text{d}t\).

Answer

\(-\int_0^3r(t)\,\text{d}t\) and \(\int_3^0r(t)\,\text{d}t\)
55016112
A continuous function \(f\) satisfies \(-2\le f(x)\le 3\) for every \(x\in[1, 5]\). a) Give the smallest guaranteed interval containing \(\int_1^5 f(x)\,\text{d}x\). b) Can \(\int_1^5 f(x)\,\text{d}x=15\)? Explain.

Hints

- Multiply each function-value bound by the interval length. - Definite integration preserves inequalities between continuous functions. - Check whether constant functions attain the two bounds. - Compare the proposed value with the resulting interval.

Solution

1. a) The interval has length \(5-1=4\). Integrating the constant bounds gives \(\int_1^5(-2)\,\text{d}x\le\int_1^5f(x)\,\text{d}x\le\int_1^5 3\,\text{d}x\). Thus, \(-8\le\int_1^5f(x)\,\text{d}x\le 12\). The constant functions \(f(x)=-2\) and \(f(x)=3\) attain the endpoints, so \([-8, 12]\) is the smallest guaranteed interval. 2. b) The value \(15\) is greater than the upper bound \(12\), so it is impossible.

Answer

a) \([-8, 12]\) b) No. The integral cannot exceed \(12\).
52456212
Justify each identity using net signed area, graph symmetry, or a horizontal shift. a) \(\int_{0}^{2\pi} \sin(x) \, \text{d}x = 0\) b) \(\int_{-2}^{2} x^4 \, \text{d}x = 2\int_{0}^{2} x^4 \, \text{d}x\) c) \(\int_{0}^{3} (x-1) \, \text{d}x = \int_{-1}^{2} x \, \text{d}x\)

Hints

- Recall the positive and negative portions of one full period of the sine curve. - Determine the symmetry of a function containing only even powers. - Consider what happens to signed area when both a graph and its interval are shifted by the same amount. - Interpret each integral as net signed area.

Solution

1. For a), the positive signed area on \([0, \pi]\) and the negative signed area on \([\pi, 2\pi]\) have equal magnitude. They cancel over one full period, so the integral is \(0\). 2. For b), \(f(x)=x^4\) is even because \(f(-x)=f(x)\). Thus the areas on \([-2, 0]\) and \([0, 2]\) are equal, so the integral over \([-2, 2]\) is twice the integral over \([0, 2]\). 3. For c), the graph of \(y=x-1\) is the graph of \(y=x\) shifted \(1\) unit to the right. The interval \([-1, 2]\) is also shifted \(1\) unit to the right to become \([0, 3]\), so the corresponding signed areas are equal.

Answer

a) The equal positive and negative signed areas over one full period of sine cancel. b) Since \(x^4\) is even, the signed areas to the left and right of the y-axis are equal. c) The graph and the interval are shifted together by \(1\) unit, so the signed area does not change.
52459312
Suppose \(f\) is integrable on \([1,3]\) and \(\int_1^3 f(x)\,\text{d}x=7\). The graph of \(h\) is the graph of \(f\) shifted \(2\) units to the right, so \(h(x)=f(x-2)\). a) Without finding an antiderivative, find \(b\) so that \(\int_3^b h(x)\,\text{d}x=7\). b) State the corresponding horizontal-translation identity for an arbitrary shift \(c\).

Hints

- Track how a horizontal shift changes both endpoints of an interval. - The accumulated signed area is preserved when the graph and interval are shifted together. - For the general statement, use the same shift amount on the function and on both bounds.

Solution

1. A shift of the graph \(2\) units to the right shifts the interval \([1,3]\) to \([3,5]\) without changing the accumulated signed area. 2. Therefore \(\int_3^5 h(x)\,\text{d}x=\int_1^3 f(x)\,\text{d}x=7\), so \(b=5\). 3. In general, if \(h(x)=f(x-c)\), then shifting both endpoints by \(c\) gives \(\int_{a+c}^{b+c} h(x)\,\text{d}x=\int_a^b f(x)\,\text{d}x\).

Answer

a) \(b=5\) b) If \(h(x)=f(x-c)\), then \(\int_{a+c}^{b+c}h(x)\,\text{d}x=\int_a^b f(x)\,\text{d}x\).
52459912
Determine whether each integral is positive, negative, or zero. Justify your answer without evaluating the integral explicitly. a) \(\int_{-5}^{5}\frac{2x}{x^2+1}\,\text{d}x\) b) \(\int_{1}^{2}(3-3^x)\,\text{d}x\) c) \(\int_{4}^{0}(x-4)\,\text{d}x\)

Hints

- Check whether an integrand has symmetry and whether its interval is symmetric about zero. - Determine the sign of each integrand on the stated interval. - Pay attention to the order of the limits of integration. - Reason from the graph's position relative to the x-axis when helpful.

Solution

1. For a), \(f(x)=\frac{2x}{x^2+1}\) is odd because \(f(-x)=-f(x)\). The interval \([-5, 5]\) is symmetric about zero, so the signed areas cancel. The integral is zero. 2. For b), \(3^x>3\) for \(x>1\), so \(3-3^x<0\) throughout \((1, 2]\), while the integrand equals zero at \(x=1\). Therefore, the integral is negative. 3. For c), \(x-4\leq 0\) on \([0, 4]\), so \(\int_{0}^{4}(x-4)\,\text{d}x\) is negative. Reversing the limits changes the sign, so \(\int_{4}^{0}(x-4)\,\text{d}x\) is positive.

Answer

a) The integral is zero. b) The integral is negative. c) The integral is positive.
52460112
Let \(f(x)=x^3-4x\). Give two different pairs of bounds \(a<b\) for which \(\int_a^b f(x)\,\text{d}x=0\). Justify one pair using symmetry and the other by calculation.

Hints

- Recall the integral property of odd functions on symmetric intervals. - For the second pair, choose one endpoint and solve for the other. - Set the antiderivative difference equal to \(0\). - Exclude the repeated-endpoint solution because \(a<b\).

Solution

1. The function is odd because \(f(-x)=-f(x)\). Therefore, its integral over any interval \([-c, c]\) is \(0\). One pair is \(a=-2\), \(b=2\). 2. For a calculated example, set \(a=0\). An antiderivative is \(F(x)=\frac{x^4}{4}-2x^2\). 3. Require \(F(b)-F(0)=0\): \(\frac{b^4}{4}-2b^2=0\). For \(b>0\), this gives \(b^2=8\), so \(b=2\sqrt{2}\). 4. A second pair is \(a=0\), \(b=2\sqrt{2}\).

Answer

One pair is \((-2, 2)\), justified by odd symmetry. A second pair is \((0, 2\sqrt{2})\), found by solving the antiderivative equation.
52461412
Each panel shows a function over the displayed interval. Without finding an antiderivative, use net signed area and graph symmetry to justify the following integral statements. a) For the blue graph \(f\), explain why \(\int_0^4 f(x)\,\text{d}x=0\). b) For the green graph \(g\), explain why \(\int_1^5 g(x)\,\text{d}x=0\). c) For the purple graph \(h\), explain why \(\int_0^{\pi} h(x)\,\text{d}x=2\int_0^{\frac{\pi}{2}} h(x)\,\text{d}x\).
Figure for problem 524614

Hints

- Inspect the x-intercepts and symmetry shown in each panel. - Identify which displayed regions contribute positive or negative signed area. - Look for either cancellation of equal-magnitude signed areas or duplication of equal positive areas.

Solution

1. In panel a), the graph forms congruent triangular regions on opposite sides of the x-axis over \([0,2]\) and \([2,4]\). Their signed areas have equal magnitude and opposite signs, so \(\int_0^4 f(x)\,\text{d}x=0\). 2. In panel b), the graph has point symmetry about \((3,0)\), and the interval \([1,5]\) is symmetric about \(x=3\). The negative and positive signed areas therefore cancel, so \(\int_1^5 g(x)\,\text{d}x=0\). 3. In panel c), the graph on \([0,\pi]\) is symmetric about \(x=\frac{\pi}{2}\). The two half-intervals contribute equal positive areas, so \(\int_0^{\pi} h(x)\,\text{d}x=2\int_0^{\frac{\pi}{2}} h(x)\,\text{d}x\).

Answer

a) The two congruent triangular regions have equal magnitude and opposite signs, so the net signed area is \(0\). b) Point symmetry about \((3,0)\) makes the signed areas on the two halves equal in magnitude and opposite in sign, so they cancel. c) Symmetry about \(x=\frac{\pi}{2}\) makes the two positive half-interval areas equal, so the full integral is twice the first half.
52463312
The blue graph \(f\) and a dashed horizontal reference line are shown. Over the displayed interval, the blue arc is an upper semicircle above the dashed line. Use the graph to write and evaluate the definite integral over the full displayed blue arc. Use geometry rather than an antiderivative, and show the rectangle and semicircle contributions separately.
Figure for problem 524633

Hints

- Read the two x-coordinates where the blue arc meets the dashed line. - Read the y-value of the dashed horizontal line from the axis. - Separate the region into a rectangle and a semicircle.

Solution

1. From the graph, the full blue arc runs from \(x=-5\) to \(x=5\), and the dashed horizontal line is \(y=3\). 2. The rectangular contribution below the dashed line has width \(10\) and height \(3\), so its area is \(30\). 3. The semicircle has diameter \(10\), so its radius is \(5\). Its area is \(\frac{1}{2}\pi(5)^2=\frac{25\pi}{2}\). 4. Therefore, \(\int_{-5}^{5}f(x)\,\text{d}x=30+\frac{25\pi}{2}\).

Answer

The graph gives width \(10\), baseline height \(3\), and semicircle radius \(5\). Thus \(\int_{-5}^{5}f(x)\,\text{d}x=10(3)+\frac{1}{2}\pi(5)^2=30+\frac{25\pi}{2}\).
52463412
The displayed blue curve \(q\) is a quarter-circle arc, and the green curve \(l\) is a line segment. Use the graph and geometry to evaluate \(\int_0^4[q(x)-l(x)]\,\text{d}x\).
Figure for problem 524634

Hints

- Read the radius of the quarter circle from the axes. - Interpret the area under the line segment as a triangle. - Use the difference rule after finding the two geometric areas.

Solution

1. From the graph, \(q\) is a quarter circle of radius \(4\), so \(\int_0^4 q(x)\,\text{d}x=\frac14\pi(4)^2=4\pi\). 2. The graph of \(l\) forms a right triangle with base \(4\) and height \(4\), so \(\int_0^4 l(x)\,\text{d}x=\frac12\cdot4\cdot4=8\). 3. By the difference rule, \(\int_0^4[q(x)-l(x)]\,\text{d}x=4\pi-8\).

Answer

\(4\pi-8\)
52470712
Use properties of definite integrals to evaluate each expression efficiently. a) \(\int_{1}^{3}(x^3+2x) \, \text{d}x-\int_{1}^{3}(x^3-4) \, \text{d}x\) b) \(\int_{0}^{\frac{\pi}{2}}\sin(x) \, \text{d}x+\int_{\frac{\pi}{2}}^{\pi}\sin(x) \, \text{d}x\) c) \(2\int_{1}^{2}\frac{1}{x} \, \text{d}x+\int_{1}^{2}(x-\frac{2}{x}) \, \text{d}x\)

Hints

- Combine integrals that have the same limits. - Adjacent intervals can be joined when the integrand is the same. - A constant factor outside an integral may be moved into the integrand. - Look for terms that cancel after the integrals are combined.

Solution

1. For a), combine the integrals because they have the same limits: \(\int_1^3[(x^3+2x)-(x^3-4)]\,\text{d}x=\int_1^3(2x+4)\,\text{d}x\). 2. Evaluate: \([x^2+4x]_1^3=(9+12)-(1+4)=16\). 3. For b), use additivity over adjacent intervals: \(\int_0^{\frac{\pi}{2}}\sin(x)\,\text{d}x+\int_{\frac{\pi}{2}}^\pi\sin(x)\,\text{d}x=\int_0^\pi\sin(x)\,\text{d}x\). 4. Evaluate: \([-\cos(x)]_0^\pi=1-(-1)=2\). 5. For c), combine the integrals over the same interval: \(\int_1^2[\frac{2}{x}+x-\frac{2}{x}]\,\text{d}x=\int_1^2x\,\text{d}x\). 6. Evaluate: \([\frac{1}{2}x^2]_1^2=2-\frac{1}{2}=\frac{3}{2}\).

Answer

a) \(16\) b) \(2\) c) \(\frac{3}{2}\)
52470812
Use properties of definite integrals to evaluate each expression efficiently. a) \(\int_{-3}^{3}[x^3+\sin(x)] \, \text{d}x\) b) \(\int_{0}^{1}(3x+1)^2 \, \text{d}x+\int_{1}^{0}(9x^2+1) \, \text{d}x\) c) \(\int_{1}^{2.5}\frac{1}{x^2} \, \text{d}x+\int_{2.5}^{5}\frac{1}{x^2} \, \text{d}x\)

Hints

- For symmetric limits, check whether the integrand is even or odd. - What happens when the limits of an integral are reversed? - Can adjacent intervals with the same integrand be combined? - Expand algebraic expressions only after combining the integrals.

Solution

1. For a), both \(x^3\) and \(\sin(x)\) are odd, so their sum is odd. The integral of an odd function over \([-3, 3]\) is \(0\). 2. For b), reverse the limits of the second integral and combine: \(\int_0^1[(3x+1)^2-(9x^2+1)]\,\text{d}x=\int_0^1 6x\,\text{d}x\). 3. Evaluate: \([3x^2]_0^1=3\). 4. For c), use additivity over adjacent intervals: \(\int_1^{2.5}\frac{1}{x^2}\,\text{d}x+\int_{2.5}^{5}\frac{1}{x^2}\,\text{d}x=\int_1^5x^{-2}\,\text{d}x\). 5. Evaluate: \([-\frac{1}{x}]_1^5=-\frac{1}{5}+1=\frac{4}{5}\).

Answer

a) \(0\) b) \(3\) c) \(\frac{4}{5}\)
52496712
Let \(f(x)=x^3-4x\). 1. Verify by evaluating the integral that \(\int_{-2}^{2}f(x) \, \text{d}x=0\). 2. Interpret the result geometrically using the terms “net signed area” and “origin symmetry.” 3. Give another function \(g\), not a constant multiple of \(f\), such that \(\int_{-a}^{a}g(x) \, \text{d}x=0\) for every \(a>0\). Briefly justify your choice.

Hints

- How do the function values change when \(x\) is replaced by \(-x\)? - Use an antiderivative to evaluate the definite integral. - Recall how regions above and below the x-axis contribute to net signed area. - Think of a familiar odd function that is not a constant multiple of \(f\).

Solution

1. An antiderivative is \(F(x)=\frac{1}{4}x^4-2x^2\). Since \(F(2)=-4\) and \(F(-2)=-4\), \(\int_{-2}^{2}(x^3-4x)\,\text{d}x=F(2)-F(-2)=0\). 2. The function is odd, so its graph has origin symmetry. On the symmetric interval \([-2, 2]\), corresponding regions above and below the x-axis have equal area and opposite signs. Their net signed area is \(0\). 3. One example is \(g(x)=\sin(x)\). It is odd and is not a constant multiple of \(f\), so its integral over every symmetric interval \([-a, a]\) is \(0\).

Answer

1. \(\int_{-2}^{2}(x^3-4x)\,\text{d}x=0\) 2. Origin symmetry makes the equal-magnitude signed areas on the two halves cancel, so the net signed area is \(0\). 3. One possible answer is \(g(x)=\sin(x)\), because it is odd and is not a constant multiple of \(f\).
52508212
The graph of \(f\) is shown. Use the graph's symmetry and the additivity of definite integrals to show that \(\int_1^3 f(x)\,\text{d}x=2\int_2^3 f(x)\,\text{d}x\).
Figure for problem 525082

Hints

- Identify the symmetry axis from the displayed graph. - Compare the intervals \([1,2]\) and \([2,3]\) across that axis. - Split the integral over \([1,3]\) at \(x=2\).

Solution

1. The displayed parabola is symmetric about \(x=2\). 2. The intervals \([1,2]\) and \([2,3]\) are mirror images across that line, so \(\int_1^2 f(x)\,\text{d}x=\int_2^3 f(x)\,\text{d}x\). 3. By additivity, \(\int_1^3 f(x)\,\text{d}x=\int_1^2 f(x)\,\text{d}x+\int_2^3 f(x)\,\text{d}x=2\int_2^3 f(x)\,\text{d}x\).

Answer

The graph is symmetric about \(x=2\), so the integrals over \([1,2]\) and \([2,3]\) are equal. Additivity then gives \(\int_1^3 f(x)\,\text{d}x=2\int_2^3 f(x)\,\text{d}x\).
52688112
Decide whether the statement is true or false: \(\int_{0}^{4}(x^3-3x^2+2x) \, \text{d}x=\int_{2}^{4}(x^3-3x^2+2x) \, \text{d}x\). Justify your decision using symmetry of \(f(x)=x^3-3x^2+2x\).

Hints

- Factor the polynomial and locate its zeros. - Test for point symmetry about the midpoint of \([0, 2]\). - What is the integral of a point-symmetric graph over an interval centered at its symmetry point?

Solution

1. Factor the function: \(f(x)=x(x-1)(x-2)\). Its graph has point symmetry about \((1, 0)\), because \(f(1+h)=-f(1-h)\). 2. The interval \([0, 2]\) is symmetric about \(x=1\). Therefore, the signed areas on its two halves cancel, so \(\int_0^2f(x)\,\text{d}x=0\). 3. By additivity, \(\int_0^4f(x)\,\text{d}x=\int_0^2f(x)\,\text{d}x+\int_2^4f(x)\,\text{d}x=\int_2^4f(x)\,\text{d}x\). 4. The statement is true.

Answer

True. Point symmetry about \((1, 0)\) makes \(\int_0^2f(x)\,\text{d}x=0\), so additivity gives \(\int_0^4f(x)\,\text{d}x=\int_2^4f(x)\,\text{d}x\).
52688212
Let \(f(x)=\cos(x)\). a) Use net signed area to explain why \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}f(x)\,\text{d}x\) is positive even though the interval is symmetric about zero. b) Find an interval \([a, b]\) with \(a>0\) such that \(\int_{a}^{b}f(x)\,\text{d}x=0\), and the graph of \(f\) lies both above and below the x-axis on the interval.

Hints

- Distinguish between even symmetry and odd symmetry when reasoning about integrals. - Determine where cosine is positive and negative. - For b), use the fact that sine is an antiderivative of cosine. - Choose endpoints with the same sine value and include a sign change of cosine.

Solution

1. For a), cosine is even, so its graph is symmetric about the y-axis. On \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), \(\cos(x)\geq 0\). Therefore, there are no negative signed-area contributions, and the integral is positive. In fact, \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(x)\,\text{d}x=\sin\left(\frac{\pi}{2}\right)-\sin\left(-\frac{\pi}{2}\right)=2\). 2. For b), \(\int_{a}^{b}\cos(x)\,\text{d}x=\sin(b)-\sin(a)\). To make the integral zero, choose endpoints with equal sine values while ensuring that cosine changes sign on the interval. 3. One valid choice is \([\pi, 2\pi]\), because \(\sin(2\pi)-\sin(\pi)=0\). On this interval, cosine is negative from \(\pi\) to \(\frac{3\pi}{2}\) and positive from \(\frac{3\pi}{2}\) to \(2\pi\).

Answer

a) The integral is positive because \(\cos(x)\geq 0\) on \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\); symmetry about the y-axis does not create negative signed area. b) One valid interval is \([\pi, 2\pi]\). On this interval, the negative and positive signed-area contributions cancel, so \(\int_{\pi}^{2\pi}\cos(x)\,\text{d}x=0\).
52688512
Two students discuss symmetry and definite integrals over intervals of the form \([-a, a]\). Maya says, “If a continuous function \(f\) is odd, then \(\int_{-a}^{a}f(x) \, \text{d}x=0\) for every \(a>0\).” Liam says, “The converse must also be true. If \(\int_{-a}^{a}f(x) \, \text{d}x=0\) for one particular \(a>0\), then the graph of \(f\) must be symmetric about the origin on \([-a, a]\).” Evaluate both claims. Disprove any false claim with a specific function and interval.

Hints

- Interpret a definite integral as net signed area. - Recall the definitions of even and odd functions. - For a counterexample, try shifting a simple even function vertically so its net signed area is \(0\). - A single zero integral does not determine the graph's symmetry.

Solution

1. Maya is correct. If \(f\) is odd, then \(f(-x)=-f(x)\), so corresponding signed areas on \([-a, 0]\) and \([0, a]\) cancel. 2. Liam is incorrect. A zero integral only says that the net signed area is \(0\); it does not force odd symmetry. 3. Consider \(f(x)=x^2-\frac{1}{3}\) on \([-1, 1]\). Then \(\int_{-1}^{1}(x^2-\frac{1}{3})\,\text{d}x=[\frac{1}{3}x^3-\frac{1}{3}x]_{-1}^{1}=0\). 4. However, \(f(-x)=f(x)\), so the function is even rather than odd. This is a counterexample to Liam's claim.

Answer

Maya is correct, and Liam is incorrect. A counterexample is \(f(x)=x^2-\frac{1}{3}\) on \([-1, 1]\): its integral is \(0\), but the function is even, not odd.
52955612
Consider \(g(x)=x^3-x\). 1) Without using an antiderivative, explain why \(I=\int_{-1}^{1}(x^3-x)\,\text{d}x=0\). 2) Use geometric reasoning to decide whether \(J=\int_{-1}^{2}(x^3-x)\,\text{d}x\) is positive or negative. 3) Confirm your conclusion about \(J\) by calculating it.

Hints

- Check the symmetry of the function and relate it to signed area on a symmetric interval. - Split the larger interval at a point where the first result can be used. - Determine the sign of the function on the remaining subinterval.

Solution

1. The function is odd because \(g(-x)=-g(x)\). On the symmetric interval \([-1, 1]\), the signed-area contributions on the left and right of the origin are equal in magnitude and opposite in sign. Therefore, \(I=0\). 2. Split the integral as \(\int_{-1}^{1}g(x)\,\text{d}x+\int_{1}^{2}g(x)\,\text{d}x\). The first integral is \(0\), and \(g(x)>0\) on \((1, 2]\). Therefore, \(J\) is positive. 3. An antiderivative is \(G(x)=\frac{1}{4}x^4-\frac{1}{2}x^2\). Thus, \(J=G(2)-G(-1)=\left(4-2\right)-\left(\frac{1}{4}-\frac{1}{2}\right)=2+\frac{1}{4}=\frac{9}{4}=2.25\).

Answer

1) \(I=0\) because an odd function has canceling signed-area contributions on an interval symmetric about zero. 2) \(J\) is positive. 3) \(J=\frac{9}{4}=2.25\).
52956612
For real \(a\), define \(I(a)=\int_0^{\frac{\pi}{2}}a\cos(x)\,\text{d}x\). a) Express \(I(a)\) in terms of \(I(1)\), and evaluate \(I(a)\). b) Determine the values of \(a\) for which \(I(a)\) is positive, zero, or negative. c) Find \(a\) if \(I(a)=-3\).

Hints

- Move the constant \(a\) outside the integral. - Evaluate the base integral \(I(1)\). - Once \(I(a)\) is known, its sign is determined by the sign of \(a\). - Translate the final condition into an equation for \(a\).

Solution

1. By the constant-multiple property, \(I(a)=aI(1)\). Since \(I(1)=\int_0^{\frac{\pi}{2}}\cos(x)\,\text{d}x=[\sin(x)]_0^{\frac{\pi}{2}}=1\), it follows that \(I(a)=a\). 2. Therefore, \(I(a)>0\) when \(a>0\), \(I(a)=0\) when \(a=0\), and \(I(a)<0\) when \(a<0\). 3. If \(I(a)=-3\), then \(a=-3\).

Answer

a) \(I(a)=aI(1)=a\) b) Positive for \(a>0\), zero for \(a=0\), and negative for \(a<0\) c) \(a=-3\)
52958312
Evaluate each integral using symmetry. 1. \(\int_{-4}^{4}(x^5-12x^3+2x) \, \text{d}x\) 2. \(\int_{-1}^{1}(15x^4-6x^2) \, \text{d}x\) 3. \(\int_{-3}^{3}(x^2+2x+1) \, \text{d}x\)

Hints

- Determine whether each integrand is even, odd, or a sum of both. - An odd function integrates to \(0\) over a symmetric interval. - For an even function, double the integral over the right half. - Use linearity to separate even and odd parts.

Solution

1. The first integrand is odd, so its integral over \([-4, 4]\) is \(0\). 2. The second integrand is even. Therefore, \(\int_{-1}^{1}(15x^4-6x^2)\,\text{d}x=2\int_0^1(15x^4-6x^2)\,\text{d}x\). An antiderivative is \(3x^5-2x^3\), so the value is \(2(3-2)=2\). 3. Split the integrand into the even part \(x^2+1\) and the odd part \(2x\). The odd part integrates to \(0\), and \(2\int_0^3(x^2+1)\,\text{d}x=2[\frac{1}{3}x^3+x]_0^3=24\).

Answer

1. \(0\) 2. \(2\) 3. \(24\)
52958412
Let \(f(x)=x^3-9x\). a) Use symmetry to explain why \(\int_{-3}^{3}f(x)\,\text{d}x=0\). b) Find \(I=\int_{-3}^{3}[f(x)+c]\,\text{d}x\) in terms of \(c\in\mathbb{R}\). c) Tereza claims, “If \(g\) is even, then \(\int_{-a}^{a}g(x)\,\text{d}x\) must be positive.” Evaluate the claim with a brief explanation or counterexample.

Hints

- Relate the powers in the polynomial to odd symmetry. - Split the sum in part b using linearity. - The sign of an integral depends on the graph's position relative to the x-axis. - Symmetry alone does not determine whether an integral is positive.

Solution

1. For part a, \(f\) is odd because it contains only odd powers. Therefore, its signed areas cancel over \([-3,3]\), and the integral is \(0\). 2. For part b, use linearity: \(I=\int_{-3}^{3}f(x)\,\text{d}x+\int_{-3}^{3}c\,\text{d}x=0+6c=6c\). 3. For part c, the claim is false. Even symmetry only makes the left and right contributions equal; it does not determine their sign. For example, \(g(x)=-1\) is even, but \(\int_{-1}^{1}-1\,\text{d}x=-2\).

Answer

a) The function is odd, so the signed areas cancel and the integral is \(0\). b) \(I=6c\) c) False. For example, \(g(x)=-1\) is even, but \(\int_{-1}^{1}g(x)\,\text{d}x=-2\).
52958812
Let \(f(x)=x^3+2x+0.5\). First combine the integrals, then evaluate \(I=\int_{-4}^{1}f(x) \, \text{d}x-\int_{4}^{1}f(x) \, \text{d}x\).

Hints

- Reverse the limits of the subtracted integral. - Combine the adjacent intervals. - After combining, identify the odd and even parts of the integrand.

Solution

1. Reverse the limits of the second integral: \(-\int_4^1f(x)\,\text{d}x=\int_1^4f(x)\,\text{d}x\). 2. Combine adjacent intervals: \(I=\int_{-4}^{4}f(x)\,\text{d}x\). 3. The terms \(x^3\) and \(2x\) are odd, so their integrals over \([-4, 4]\) are \(0\). 4. Only the constant remains: \(I=\int_{-4}^{4}0.5\,\text{d}x=0.5\cdot 8=4\).

Answer

\(4\)
52958912
The graph of \(f\) is symmetric about the y-axis. Suppose \(\int_{0}^{3}f(x) \, \text{d}x=4\) and \(\int_{3}^{5}f(x) \, \text{d}x=-2\). Use symmetry and properties of definite integrals to find each value. a) \(\int_{-3}^{3}f(x) \, \text{d}x\) b) \(\int_{-5}^{-3}f(x) \, \text{d}x\) c) \(\int_{-5}^{0}f(x) \, \text{d}x\) d) \(\int_{0}^{5}[f(x)+3] \, \text{d}x\)

Hints

- Mirror intervals have equal integrals for an even function. - Use additivity to combine \([0, 3]\) and \([3, 5]\). - Split the sum in part d) using linearity. - Integrate the constant over the full interval.

Solution

1. For a), even symmetry gives \(\int_{-3}^{3}f(x)\,\text{d}x=2\int_0^3f(x)\,\text{d}x=8\). 2. For b), mirror intervals have equal integrals for an even function: \(\int_{-5}^{-3}f(x)\,\text{d}x=\int_3^5f(x)\,\text{d}x=-2\). 3. For c), \(\int_{-5}^{0}f(x)\,\text{d}x=\int_0^5f(x)\,\text{d}x=4+(-2)=2\). 4. For d), use linearity: \(\int_0^5[f(x)+3]\,\text{d}x=\int_0^5f(x)\,\text{d}x+\int_0^5 3\,\text{d}x=2+15=17\).

Answer

a) \(8\) b) \(-2\) c) \(2\) d) \(17\)
52959412
Let \(f(x)=x^3-6x^2+2x\). a) Show that, for \(k>0\), the value of \(I=\int_{-k}^{k}f(x) \, \text{d}x\) is determined entirely by the quadratic term. b) Find \(I\) in terms of \(k\).

Hints

- Separate the polynomial into even and odd parts. - Odd terms integrate to \(0\) over symmetric intervals. - Use even symmetry to evaluate the remaining quadratic term.

Solution

1. Use linearity to separate the odd and even parts: \(f(x)=(x^3+2x)-6x^2\). 2. The function \(x^3+2x\) is odd, so its integral over \([-k, k]\) is \(0\). Therefore, only the even term \(-6x^2\) contributes. 3. Compute: \(I=2\int_0^k-6x^2\,\text{d}x=2[-2x^3]_0^k=-4k^3\).

Answer

a) The odd terms \(x^3\) and \(2x\) integrate to \(0\) over \([-k, k]\), leaving only \(-6x^2\). b) \(I=-4k^3\)
52961112
Suman claims that \(\int_{-2}^{2}(2x^5-4x^3+3x^2-1)\,\text{d}x=2\int_0^2(2x^5-4x^3+3x^2-1)\,\text{d}x\) because the limits are symmetric about \(0\). Explain why this reasoning is incorrect, then evaluate the original integral efficiently.

Hints

- Check whether the entire integrand is even before doubling a half-interval integral. - Separate the polynomial into its odd and even parts. - The odd part contributes \(0\) over symmetric limits. - Double only the integral of the even part.

Solution

1. Doubling the integral over \([0,2]\) is valid for an even integrand, but the given integrand contains both odd and even terms. Therefore, Suman's equation is not justified. 2. Separate the odd part \(2x^5-4x^3\) from the even part \(3x^2-1\). The odd part integrates to \(0\) over \([-2,2]\). 3. Thus, the integral equals \(2\int_0^2(3x^2-1)\,\text{d}x\). An antiderivative is \(x^3-x\), so the value is \(2[x^3-x]_0^2=2(8-2)=12\).

Answer

Suman's rule applies only to an even integrand. After removing the odd part by symmetry, \(\int_{-2}^{2}(2x^5-4x^3+3x^2-1)\,\text{d}x=2\int_0^2(3x^2-1)\,\text{d}x=12\).
52968512
Prove the constant multiple rule for definite integrals: \(\int_a^b kf(x)\,\text{d}x=k\int_a^b f(x)\,\text{d}x\), where \(k\in\mathbb{R}\). Use the Fundamental Theorem of Calculus and the corresponding derivative rule.

Hints

- Use the derivative rule for a constant multiple. - Identify an antiderivative of \(kf\) from an antiderivative of \(f\). - Apply the Fundamental Theorem at the endpoints. - Factor \(k\) from the resulting difference.

Solution

1. Let \(F\) be an antiderivative of \(f\), so \(F'(x)=f(x)\). 2. By the constant multiple rule for derivatives, \([kF(x)]'=kF'(x)=kf(x)\). Thus \(kF\) is an antiderivative of \(kf\). 3. By the Fundamental Theorem of Calculus, \(\int_a^b kf(x)\,\text{d}x=[kF(x)]_a^b=kF(b)-kF(a)\). 4. Factor out \(k\): \(k[F(b)-F(a)]=k\int_a^b f(x)\,\text{d}x\).

Answer

If \(F'=f\), then \((kF)'=kf\). Therefore, the Fundamental Theorem gives \(\int_a^b kf(x)\,\text{d}x=[kF(x)]_a^b=k[F(b)-F(a)]=k\int_a^b f(x)\,\text{d}x\).
52968612
Let \(f\) and \(g\) be continuous on \([a, b]\). Use the Fundamental Theorem of Calculus to prove \(\int_a^b[f(x)-g(x)]\,\text{d}x=\int_a^b f(x)\,\text{d}x-\int_a^b g(x)\,\text{d}x\).

Hints

- Differentiate the difference of two antiderivatives. - Use the derivative difference rule. - Apply the Fundamental Theorem to the difference. - Regroup the endpoint terms into two separate differences.

Solution

1. Let \(F'=f\) and \(G'=g\). By the difference rule for derivatives, \((F-G)'=F'-G'=f-g\). 2. Thus \(F-G\) is an antiderivative of \(f-g\). By the Fundamental Theorem, \(\int_a^b[f(x)-g(x)]\,\text{d}x=[F(x)-G(x)]_a^b\). 3. Expand the endpoint expression: \([F(b)-G(b)]-[F(a)-G(a)]=[F(b)-F(a)]-[G(b)-G(a)]\). 4. Applying the Fundamental Theorem to each difference gives the desired result.

Answer

Because \((F-G)'=f-g\), the Fundamental Theorem gives \(\int_a^b(f-g)\,\text{d}x=[F-G]_a^b=[F(b)-F(a)]-[G(b)-G(a)]=\int_a^b f\,\text{d}x-\int_a^b g\,\text{d}x\).
52969412
Find the real constant \(c\) such that \(\int_{-2}^{2}\left(\frac{1}{2}x^3-x^2+c\right)\,\text{d}x=0\). Use symmetry, and explain why the cubic term does not affect the value of \(c\).

Hints

- Separate the odd and even parts of the integrand. - The odd cubic term vanishes over \([-2, 2]\). - Double the integral of the even part over \([0, 2]\). - Set the resulting expression equal to \(0\) and solve for \(c\).

Solution

1. The term \(\frac{1}{2}x^3\) is odd, so its integral over \([-2, 2]\) is \(0\). Therefore, the cubic term does not affect the condition. 2. The remaining integrand \(-x^2+c\) is even, so the condition becomes \(2\int_0^2(-x^2+c)\,\text{d}x=0\). 3. Evaluate: \(2[-\frac{1}{3}x^3+cx]_0^2=2(-\frac{8}{3}+2c)=-\frac{16}{3}+4c\). 4. Set this equal to \(0\): \(4c=\frac{16}{3}\), so \(c=\frac{4}{3}\).

Answer

The cubic term is odd and contributes \(0\). The remaining condition gives \(c=\frac{4}{3}\).
53269912
The graph shown encloses three regions with the x-axis on \([-3, 5]\). Their areas, rounded to two decimal places, are \(A_1=1.50\) on \([-3, -1]\), \(A_2=2.75\) on \([-1, 2]\), and \(A_3=5.35\) on \([2, 5]\). Use these values to evaluate each integral. a) \(\int_{-3}^{2}f(x)\,\text{d}x\) b) \(\int_{-1}^{5}f(x)\,\text{d}x\) c) \(\int_{5}^{-3}f(x)\,\text{d}x\)
Figure for problem 532699

Hints

- Determine the sign of each region from its position relative to the x-axis. - Split each integral into intervals matching the labeled areas. - Reversing the limits of integration changes the sign of the integral.

Solution

1. For a), the first region is below the x-axis and the second is above it. Thus, \(\int_{-3}^{2}f(x)\,\text{d}x=-A_1+A_2=-1.50+2.75=1.25\). 2. For b), the second region is above the axis and the third is below it. Thus, \(\int_{-1}^{5}f(x)\,\text{d}x=A_2-A_3=2.75-5.35=-2.60\). 3. For c), first find the integral in the forward direction: \(\int_{-3}^{5}f(x)\,\text{d}x=-A_1+A_2-A_3=-1.50+2.75-5.35=-4.10\). Reversing the limits changes the sign, so \(\int_{5}^{-3}f(x)\,\text{d}x=4.10\).

Answer

a) \(\int_{-3}^{2}f(x)\,\text{d}x=1.25\) b) \(\int_{-1}^{5}f(x)\,\text{d}x=-2.60\) c) \(\int_{5}^{-3}f(x)\,\text{d}x=4.10\)
53270212
Let \(f(x)=0.5x^3-3x^2+4x\). The function is symmetric about the point \(P(2,0)\). a) Use the symmetry to explain why \(\int_0^4f(x)\,\text{d}x=0\). b) Find the total area between the graph and the x-axis on \([0,4]\). c) Define \(g(x)=f(x)+1\). Without finding an antiderivative of \(g\), use properties of definite integrals to explain why \(\int_0^4g(x)\,\text{d}x=4\).

Hints

- Relate point symmetry about a point on the x-axis to signed areas on opposite sides of that point. - Distinguish net signed area from total geometric area. - Use the sum rule for definite integrals in part c).

Solution

1. Point symmetry about \((2,0)\) makes the region above the x-axis on \([0,2]\) congruent to the region below the axis on \([2,4]\). Their signed contributions cancel, so \(\int_0^4f(x)\,\text{d}x=0\). 2. An antiderivative is \(F(x)=\frac18x^4-x^3+2x^2\). The positive contribution on \([0,2]\) is \(F(2)-F(0)=2\). By symmetry, the region on \([2,4]\) also has geometric area \(2\). The total area is \(4\) square units. 3. By linearity, \(\int_0^4g(x)\,\text{d}x=\int_0^4f(x)\,\text{d}x+\int_0^4 1\,\text{d}x=0+4=4\).

Answer

a) The equal-magnitude positive and negative signed areas cancel, so \(\int_0^4f(x)\,\text{d}x=0\). b) \(4\) square units c) \(\int_0^4g(x)\,\text{d}x=4\).
53462412
The graph of \(f\) is shown. Determine whether each statement is true or false, and justify your answer from the graph. a) \(\int_{-1}^{1}f(x)\,\text{d}x<0\) b) \(\int_{-1}^{3}f(x)\,\text{d}x=0\) c) \(\int_{0}^{2}f(x)\,\text{d}x=0\) d) \(\int_{3}^{4}f(x)\,\text{d}x>0\)
Figure for problem 534624

Hints

- Use the sign of the function to determine the sign of an integral. - Look for point symmetry about a point on the x-axis. - Equal regions on opposite sides of the axis can cancel in a signed-area calculation. - An integral is negative when the function is negative throughout the interval.

Solution

1. The graph is below the x-axis on \((-1, 1)\), above it on \((1, 3)\), and below it again for \(x>3\). 2. For a), the function is negative throughout the interior of \([-1, 1]\), so the integral is negative. The statement is true. 3. For b), the graph has point symmetry about \((1, 0)\). The regions on \([-1, 1]\) and \([1, 3]\) have equal area and opposite signs, so the integral is zero. The statement is true. 4. For c), the interval \([0, 2]\) is also symmetric about \(x=1\). The negative contribution on \([0, 1]\) cancels the equal positive contribution on \([1, 2]\). The statement is true. 5. For d), the graph is below the x-axis on \((3, 4)\), so the integral is negative. The statement is false.

Answer

a) True b) True c) True d) False
53462912
The graph shows three labeled regions between \(f\) and the x-axis with areas \(A_1=1.2\), \(A_2=3.5\), and \(A_3=0.8\). Use these values to evaluate the integrals. a) \(\int_{-4}^{-1}f(x)\,\text{d}x\) b) \(\int_{-1}^{5}f(x)\,\text{d}x\) c) \(\int_{-4}^{5}f(x)\,\text{d}x\) d) \(\int_{3}^{-1}f(x)\,\text{d}x\)
Figure for problem 534629

Hints

- Determine which labeled regions lie above and below the x-axis. - Combine the signed contributions of all regions within each interval. - Reversing the limits of integration changes the sign.

Solution

1. Region \(A_1\) lies above the x-axis, so \(\int_{-4}^{-1}f(x)\,\text{d}x=A_1=1.2\). 2. On \([-1, 5]\), region \(A_2\) lies below the axis and \(A_3\) lies above it. Therefore, \(\int_{-1}^{5}f(x)\,\text{d}x=-A_2+A_3=-3.5+0.8=-2.7\). 3. On \([-4, 5]\), all three regions contribute: \(\int_{-4}^{5}f(x)\,\text{d}x=A_1-A_2+A_3=1.2-3.5+0.8=-1.5\). 4. Reversing limits changes the sign. Since \(\int_{-1}^{3}f(x)\,\text{d}x=-A_2=-3.5\), \(\int_{3}^{-1}f(x)\,\text{d}x=3.5\).

Answer

a) \(1.2\) b) \(-2.7\) c) \(-1.5\) d) \(3.5\)
53465812
The figure shows the graphs of \(f\) and \(g\). Use the figure and the geometric meaning of a definite integral to explain why \(\int_{-1}^{1}f(x)\,\text{d}x=\int_{1}^{3}g(x)\,\text{d}x\).
Figure for problem 534658

Hints

- Compare the location and shape of the two displayed curves. - Compare the interval endpoints marked in the figure. - Decide whether a horizontal translation changes the area of the corresponding region.

Solution

1. The displayed graph of \(g\) is the graph of \(f\) shifted \(2\) units to the right. 2. The interval \([1,3]\) is also \([-1,1]\) shifted \(2\) units to the right. 3. A horizontal translation of both a graph and its interval preserves the area between the graph and the x-axis. Therefore the two definite integrals are equal.

Answer

The figure shows that \(g\) is a horizontal translation of \(f\) by \(2\) units, and \([1,3]\) is the same translation of \([-1,1]\). The corresponding regions have equal area, so the integrals are equal.
53467612
The graph shown is an even quadratic polynomial \(h\). 1) Find the formula for \(h\) using the zeros and the y-intercept shown. 2) Verify algebraically that the graph is symmetric about the y-axis. 3) Without using an antiderivative, explain why \(\int_{-3}^{3}h(x)\,\text{d}x=2\int_{0}^{3}h(x)\,\text{d}x\). 4) Evaluate \(\int_{-3}^{3}h(x)\,\text{d}x\) and interpret the result as area between the graph and the x-axis.
Figure for problem 534676

Hints

- Use the zeros to write a factored quadratic formula. - Use the y-intercept to find the leading coefficient. - Check the condition \(h(-x)=h(x)\). - Use symmetry to compare the integrals on the left and right halves.

Solution

1. The zeros are \(x=-3\) and \(x=3\), and the y-intercept is \(3\). Write \(h(x)=a(x+3)(x-3)=a(x^2-9)\). Since \(h(0)=3\), \(-9a=3\), so \(a=-\frac{1}{3}\). Therefore, \(h(x)=-\frac{1}{3}x^2+3\). 2. \(h(-x)=-\frac{1}{3}(-x)^2+3=-\frac{1}{3}x^2+3=h(x)\). Thus, \(h\) is even and its graph is symmetric about the y-axis. 3. The regions on \([-3, 0]\) and \([0, 3]\) are congruent and both lie above the x-axis. Therefore, the full integral is twice the integral on the right half. 4. An antiderivative is \(H(x)=-\frac{1}{9}x^3+3x\). Thus, \(\int_{-3}^{3}h(x)\,\text{d}x=H(3)-H(-3)=6-(-6)=12\). Since the graph lies above the x-axis on \([-3, 3]\), this value is also the geometric area.

Answer

1) \(h(x)=-\frac{1}{3}x^2+3\) 2) \(h(-x)=h(x)\), so the graph is symmetric about the y-axis. 3) The two halves are congruent and both contribute positively. 4) \(\int_{-3}^{3}h(x)\,\text{d}x=12\), which is the area between the graph and the x-axis.
53469012
The graph of \(f(x)=0.25x^3-4x\) has point symmetry about the origin. The regions enclosed with the x-axis are labeled \(A_1\) and \(A_2\). Determine whether each statement is correct. Correct any false statement. (1) \(\int_{-4}^{4}f(x)\,\text{d}x=0\). (2) The area of \(A_2\) is \(\int_{0}^{4}f(x)\,\text{d}x\). (3) \(A_1=A_2\). (4) The total area is \(A_{\text{total}}=2\int_{-4}^{0}f(x)\,\text{d}x\).
Figure for problem 534690

Hints

- Use point symmetry to compare function values and regions on opposite sides of the origin. - Distinguish between a signed integral and a positive geometric area. - Find where the graph lies above and below the x-axis. - Check whether a total-area expression adds magnitudes rather than signed contributions.

Solution

1. Statement (1) is correct. The function is odd, and the interval is symmetric about zero, so the signed-area contributions cancel. 2. Statement (2) is false. The graph lies below the x-axis on \([0, 4]\), so \(\int_{0}^{4}f(x)\,\text{d}x=-16\). The area is positive, so the corrected statement is \(A_2=-\int_{0}^{4}f(x)\,\text{d}x\). 3. Statement (3) is correct. Point symmetry maps the two regions to each other, so they have equal area. 4. Statement (4) is correct. The graph lies above the x-axis on \([-4, 0]\), so that integral equals \(A_1\). Since \(A_1=A_2\), doubling it gives the total area.

Answer

(1) Correct (2) False; \(A_2=-\int_{0}^{4}f(x)\,\text{d}x\) (3) Correct (4) Correct
54903912
The net rate \(m(t)\), in kilograms per hour, describes the change in material on a recycling conveyor. Measurements show \(\int_0^{10}m(t)\,\text{d}t=-42\), \(\int_0^3m(t)\,\text{d}t=18\), and \(\int_7^{10}m(t)\,\text{d}t=-25\). Find \(\int_3^7m(t)\,\text{d}t\) and interpret its value.

Hints

- Break the total accumulated change into changes over adjacent time intervals. - Keep the negative sign attached to a net loss. - After finding the missing value, translate its units and sign back into the setting.

Solution

1. Split the full interval into adjacent parts: \(\int_0^{10}m(t)\,\text{d}t=\int_0^3m(t)\,\text{d}t+\int_3^7m(t)\,\text{d}t+\int_7^{10}m(t)\,\text{d}t\). 2. Substitute the known values: \(-42=18+\int_3^7m(t)\,\text{d}t-25\). 3. Solving gives \(\int_3^7m(t)\,\text{d}t=-35\). 4. The conveyor's material amount has a net decrease of \(35\) kilograms from hour \(3\) to hour \(7\).

Answer

\(\int_3^7m(t)\,\text{d}t=-35\). The material on the conveyor decreases by a net \(35\,\text{kg}\) during that interval.
54913912
Evaluate \(\int_0^4\frac{x-2}{x^2-4x+8}\,\text{d}x\) without finding an antiderivative. Complete the square, shift to a centered variable, and use symmetry. Your answer must include the transformed symmetric integral and identify its parity.

Hints

- Recenter the quadratic denominator at its vertex. - Shift the bounds using the same change of variable. - Check what happens to the transformed integrand when \(u\) is replaced by \(-u\).

Solution

1. Complete the square: \(x^2-4x+8=(x-2)^2+4\). 2. Let \(u=x-2\). The bounds become \(-2\) and \(2\), and \(\text{d}u=\text{d}x\). 3. The integral becomes \(\int_{-2}^{2}\frac{u}{u^2+4}\,\text{d}u\). 4. The transformed integrand is odd, so its integral over the symmetric interval is \(0\).

Answer

\(\int_{-2}^{2}\frac{u}{u^2+4}\,\text{d}u=0\) after \(u=x-2\). The integrand \(\frac{u}{u^2+4}\) is odd.
52496812
A continuous function \(f\) satisfies \(\int_{-k}^{k}f(x) \, \text{d}x=2\int_{0}^{k}f(x) \, \text{d}x\) for every \(k>0\). 1. Verify the equation algebraically for \(f(x)=3x^2+5\). 2. Explain which symmetry property a continuous function must have for the equation to hold for every \(k>0\). 3. Find \(c\in\mathbb{R}\) so that \(h(x)=x^2+cx+1\) satisfies the equation for every \(k>0\).

Hints

- Evaluate both sides for the given polynomial. - Rewrite the equation as equality of the integrals over the left and right halves. - Differentiate that equality with respect to \(k\). - For the polynomial in part 3, identify which term prevents even symmetry.

Solution

1. For \(f(x)=3x^2+5\), \(\int_{-k}^k(3x^2+5)\,\text{d}x=[x^3+5x]_{-k}^k=2k^3+10k\). Also, \(2\int_0^k(3x^2+5)\,\text{d}x=2[x^3+5x]_0^k=2k^3+10k\). 2. The equation is equivalent to \(\int_{-k}^{0}f(x)\,\text{d}x=\int_0^k f(x)\,\text{d}x\) for every \(k>0\). Differentiating both sides with respect to \(k\) gives \(f(-k)=f(k)\). Thus \(f\) must be even, so its graph is symmetric about the y-axis. 3. The polynomial \(h(x)=x^2+cx+1\) is even only when the odd-power term is absent. Therefore, \(c=0\). Equivalently, direct integration gives a difference of \(ck^2\) between the two sides, which vanishes for every \(k>0\) only when \(c=0\).

Answer

1. Both sides equal \(2k^3+10k\). 2. The function must be even: \(f(-x)=f(x)\). Its graph is symmetric about the y-axis. 3. \(c=0\)
52688612
Consider the family \(f_k(x)=x^4+kx^2\), where \(k\) is real. 1) Show that every function in the family satisfies \(f_k(0)=0\). 2) Find \(k\) so that \(\int_{-2}^{2}f_k(x)\,\text{d}x=0\). 3) Analyze the symmetry of the graph for the value of \(k\) from part 2. 4) Explain why this is a counterexample to the claim: “If a continuous function satisfies \(\int_{-a}^{a}f(x)\,\text{d}x=0\) and \(f(0)=0\), then its graph has origin symmetry.”

Hints

- Substitute \(x=0\). - Evaluate the integral in terms of \(k\), then set it equal to \(0\). - Test \(f(-x)\) to determine symmetry. - A counterexample satisfies the hypotheses but violates the conclusion.

Solution

1. Substitution gives \(f_k(0)=0^4+k\cdot 0^2=0\). 2. Compute the integral: \(\int_{-2}^{2}(x^4+kx^2)\,\text{d}x=\frac{64}{5}+\frac{16k}{3}\). Set this equal to \(0\): \(\frac{64}{5}+\frac{16k}{3}=0\), so \(k=-\frac{12}{5}\). 3. For this value, \(f(x)=x^4-\frac{12}{5}x^2\). Since \(f(-x)=f(x)\), the graph has y-axis symmetry, not origin symmetry. 4. This continuous, nonzero function satisfies both hypotheses but not the conclusion, so it disproves the claim.

Answer

1) \(f_k(0)=0\) 2) \(k=-\frac{12}{5}\) 3) The graph has y-axis symmetry because \(f(-x)=f(x)\). 4) The function is a counterexample because the integral is \(0\) and \(f(0)=0\), yet the graph does not have origin symmetry.
52957612
Find the real number \(k\) for which the equation \(\int_4^1[kg(x)+2] \, \text{d}x+\int_7^7[g(x)+k] \, \text{d}x+\int_1^4[3g(x)-k] \, \text{d}x=-15\) holds for every continuous function \(g\).

Hints

- An integral with identical limits is \(0\). - Rewrite all nonzero integrals using the same limits. - If the equation must hold for every \(g\), what must happen to its coefficient? - Verify the remaining constant integral.

Solution

1. The middle integral is \(0\) because its limits are equal. 2. Reverse the limits of the first integral and combine the remaining integrals over \([1, 4]\): \(\int_1^4[(3-k)g(x)-(k+2)]\,\text{d}x\). 3. For the value to be independent of the choice of \(g\), the coefficient of \(g(x)\) must be \(0\). Thus \(3-k=0\), so \(k=3\). 4. Check the constant part: \(\int_1^4[-(3+2)]\,\text{d}x=\int_1^4-5\,\text{d}x=-15\). Therefore, \(k=3\) works.

Answer

\(k=3\)
52959012
The graph of a continuous function \(g\) is symmetric about the origin, and \(\int_{-2}^{6}g(x) \, \text{d}x=12\). Find each integral if it is determined by the information given. Otherwise, state that it cannot be determined. a) \(\int_{2}^{6}g(x) \, \text{d}x\) b) \(\int_{-6}^{-2}g(x) \, \text{d}x\) c) \(\int_{0}^{6}g(x) \, \text{d}x-\int_{0}^{2}g(x) \, \text{d}x\) d) \(\int_{-6}^{6}g(x) \, \text{d}x\) e) \(\int_{0}^{2}g(x) \, \text{d}x\)

Hints

- An odd function integrates to \(0\) over every symmetric interval. - Split the given interval at \(-2\), \(0\), or \(2\) as useful. - Integrals over \([a, b]\) and \([-b, -a]\) have opposite signs for an odd function. - Check whether the information determines individual integrals or only their difference.

Solution

1. Since \(g\) is odd, \(\int_{-a}^{a}g(x)\,\text{d}x=0\) for every \(a>0\). 2. For a), split the given integral: \(12=\int_{-2}^{2}g(x)\,\text{d}x+\int_2^6g(x)\,\text{d}x=0+\int_2^6g(x)\,\text{d}x\). Thus the value is \(12\). 3. For b), origin symmetry gives \(\int_{-6}^{-2}g(x)\,\text{d}x=-\int_2^6g(x)\,\text{d}x=-12\). 4. For c), additivity gives \(\int_0^6g(x)\,\text{d}x-\int_0^2g(x)\,\text{d}x=\int_2^6g(x)\,\text{d}x=12\). 5. For d), the integral over \([-6, 6]\) is \(0\). 6. For e), the given information fixes only the difference between \(\int_0^6g(x)\,\text{d}x\) and \(\int_0^2g(x)\,\text{d}x\), not either value separately. It cannot be determined.

Answer

a) \(12\) b) \(-12\) c) \(12\) d) \(0\) e) Cannot be determined
53467512
The graph shown is a cubic polynomial \(f\). 1) Find the formula for \(f\). All zeros are integers, and the point \(P(1, -1)\) lies on the graph. 2) Verify algebraically that the graph of \(f\) has point symmetry about the origin. 3) Without calculating an antiderivative, explain why \(\int_{-2.5}^{2.5}f(x)\,\text{d}x=0\). 4) Use symmetry to explain why \(\int_{-1}^{2}f(x)\,\text{d}x=\int_{1}^{2}f(x)\,\text{d}x\).
Figure for problem 534675

Hints

- Read the integer zeros from the graph and use a factored polynomial form. - Substitute the given point to determine the leading coefficient. - Check the condition \(f(-x)=-f(x)\) for point symmetry about the origin. - Use interval additivity and the integral of an odd function over a symmetric interval.

Solution

1. The graph has zeros at \(x=-2\), \(x=0\), and \(x=2\), so \(f(x)=ax(x+2)(x-2)=a(x^3-4x)\). Since \(P(1, -1)\) is on the graph, \(-1=a(1-4)=-3a\), so \(a=\frac{1}{3}\). Therefore, \(f(x)=\frac{1}{3}x^3-\frac{4}{3}x\). 2. \(f(-x)=\frac{1}{3}(-x)^3-\frac{4}{3}(-x)=-\frac{1}{3}x^3+\frac{4}{3}x=-f(x)\). Thus, \(f\) is odd and its graph has point symmetry about the origin. 3. The interval \([-2.5, 2.5]\) is symmetric about zero. For an odd function, the signed-area contributions on the two halves are equal in magnitude and opposite in sign, so the integral is \(0\). 4. By interval additivity, \(\int_{-1}^{2}f(x)\,\text{d}x=\int_{-1}^{1}f(x)\,\text{d}x+\int_{1}^{2}f(x)\,\text{d}x\). The first integral is \(0\) because \(f\) is odd and \([-1, 1]\) is symmetric about zero. Therefore, \(\int_{-1}^{2}f(x)\,\text{d}x=\int_{1}^{2}f(x)\,\text{d}x\).

Answer

1) \(f(x)=\frac{1}{3}x^3-\frac{4}{3}x\) 2) \(f(-x)=-f(x)\), so the graph has point symmetry about the origin. 3) \(\int_{-2.5}^{2.5}f(x)\,\text{d}x=0\). 4) Since \(\int_{-1}^{1}f(x)\,\text{d}x=0\), \(\int_{-1}^{2}f(x)\,\text{d}x=\int_{1}^{2}f(x)\,\text{d}x\).

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