Free Math Worksheets

- 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.

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: \(\int_a^b f(x)\,dx=-\int_b^a f(x)\,dx\). Therefore, \(\int_1^4 f(x)\,dx+\int_4^1 f(x)\,dx=\int_1^4 f(x)\,dx-\int_1^4 f(x)\,dx=0\). 3. For c), a definite integral whose lower and upper limits are equal is always \(0\): \(\int_a^a h(x)\,dx=0\).

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 \(0\).

- Find an antiderivative using the chain rule in reverse. - Choose bounds where the antiderivative has equal values. - Use zeros of the sine function. - A full period also gives a zero integral.

1. An antiderivative is \(G(x)=\sin(2x)\). 2. Choose \(a=0\). Then the integral is zero when \(\sin(2b)-\sin(0)=0\). 3. Taking \(2b=\pi\) gives \(b=\frac{\pi}{2}\). 4. Verify: \(\int_0^{\frac{\pi}{2}}2\cos(2x)\,dx=[\sin(2x)]_0^{\frac{\pi}{2}}=\sin(\pi)-\sin(0)=0\).

One possible interval is \(\left[0, \frac{\pi}{2}\right]\).

- 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.

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 \(0\).

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 \(0\).

- Identify the vertex of the parabola. - What is the axis of symmetry? - Compare the locations and widths of the two intervals. - How does graph symmetry affect the two areas?

1. The graph of \(f(x)=(x-2)^2\) is a parabola with vertex \((2, 0)\) and axis of symmetry \(x=2\). 2. The intervals \([1, 2]\) and \([2, 3]\) have the same width and lie on opposite sides of the axis of symmetry. 3. Corresponding function values are equal on the two intervals, so the areas under the graph are equal. Therefore, the integrals are equal.

The graph is symmetric about \(x=2\), and the intervals \([1, 2]\) and \([2, 3]\) are mirror images across that line. Therefore, the two integrals are equal.

- 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.

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: \(\int_e^1\frac{1}{x}\,dx=-\int_1^e\frac{1}{x}\,dx\). Their sum is \(0\). 3. For c), linearity gives \(\int_0^2(x^2+4)\,dx=\int_0^2x^2\,dx+\int_0^2 4\,dx\). The constant integral is the area of a rectangle with width \(2\) and height \(4\), so it equals \(8\).

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 \(\int_0^2 4\,dx=8\).

- 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.

1. Apply linearity: \(\int_2^6[1.5f(x)-4g(x)+3]\,dx=1.5\int_2^6f(x)\,dx-4\int_2^6g(x)\,dx+\int_2^6 3\,dx\). 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 \(10\).

\(10\)

- 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.

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

\(c=A\)

- Use the power rule for each integral. - Factor the common expression \(k^4-1\). - Use \(k>1\) to justify canceling the common factor.

1. Evaluate \(A\): \(A=\left[\frac{x^4}{4}\right]_1^k=\frac{k^4-1}{4}\). 2. Evaluate \(B\): \(B=\left[\frac{x^4}{8}\right]_1^k=\frac{k^4-1}{8}\). 3. Since \(k>1\), \(k^4-1\ne 0\). Therefore, \(\frac{B}{A}=\frac{(k^4-1)/8}{(k^4-1)/4}=\frac{1}{2}\).

\(A=\frac{k^4-1}{4}\), \(B=\frac{k^4-1}{8}\), and \(\frac{B}{A}=\frac{1}{2}\).

- Recall an antiderivative of cosine. - Treat \(a\) as a constant. - Use the constant-multiple property of definite integrals.

1. Since an antiderivative of \(\cos(x)\) is \(\sin(x)\), \(I_1=[\sin(x)]_0^{\frac{\pi}{2}}=1\). 2. By the constant-multiple property, \(I_2=a\int_0^{\frac{\pi}{2}}\cos(x)\,dx=a\). 3. Therefore, \(I_2=aI_1\).

\(I_1=1\), \(I_2=a\), and \(I_2=aI_1\).

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

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

\(I=-16\)

- 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.

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

\(9\)

- 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.

1. Reverse the limits in the middle term: \(-\int_4^2f(x)\,dx=\int_2^4f(x)\,dx\), 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 \(\int_{-1}^{5}(3x^2-4x+5)\,dx\). 3. An antiderivative is \(F(x)=x^3-2x^2+5x\). 4. Evaluate: \(F(5)-F(-1)=100-(-8)=108\).

\(\int_{-1}^{5}(3x^2-4x+5) \, dx=108\)

- 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.

1. Combine the integrals over the same interval: \(\int_1^4\left[3\left(\sqrt{x}+\frac{1}{x^2}\right)+2-3\sqrt{x}\right]dx\). 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: \(F(4)-F(1)=\frac{29}{4}-(-1)=\frac{33}{4}\).

\(\int_{1}^{4}\left(\frac{3}{x^2}+2\right) \, dx=\frac{33}{4}\)

- 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.

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

a) \(0\) b) \(\frac{1}{10}k^5-4k\)

- Identify which terms are odd and which are even. - Odd terms vanish over a symmetric interval. - Use even symmetry to reduce the remaining computation to \([0, 5]\).

1. The terms \(x^3\) and \(-\frac{1}{2}x\) are odd, so their integrals over \([-5, 5]\) are \(0\). 2. Only the even term remains: \(\int_{-5}^{5}\frac{3}{10}x^4\,dx=2\int_0^5\frac{3}{10}x^4\,dx\). 3. An antiderivative is \(\frac{3}{50}x^5\). Therefore, \(2[\frac{3}{50}x^5]_0^5=375\).

\(375\)

- 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.

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

a) \(0\) b) \(-32\)

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

1. Apply the constant multiple rule: \(\int_1^2 5x^4\,dx=5\int_1^2x^4\,dx\). 2. An antiderivative of \(x^4\) is \(\frac15x^5\), so \(5\left[\frac15x^5\right]_1^2=[x^5]_1^2\). 3. Evaluate the bounds: \(2^5-1^5=32-1=31\).

\(31\)

- You do not need a formula for \(g\). - A constant factor can be moved outside a definite integral. - Multiply the given integral value by \(\frac23\).

1. By the constant multiple rule, \(\int_2^6\frac23g(x)\,dx=\frac23\int_2^6g(x)\,dx\). 2. Substitute the given value: \(\frac23\cdot18=12\).

\(12\), because \(\int_2^6\frac23g(x)\,dx=\frac23\int_2^6g(x)\,dx\).

- Split the integral into two simpler integrals. - Rewrite \(\frac1{x^2}\) using a negative exponent. - After finding an antiderivative, evaluate it at the upper and lower bounds.

1. Split the integral and rewrite the reciprocal power: \(\int_1^2 6x^2\,dx-4\int_1^2x^{-2}\,dx\). 2. An antiderivative of the integrand is \(F(x)=2x^3+\frac4x\). 3. Apply the fundamental theorem of calculus: \(F(2)-F(1)=\left(16+2\right)-\left(2+4\right)=18-6=12\).

\(12\)

- 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.

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 \(\int_{-4}^{-1}f(x)\,dx=-A_1=-4\). 3. For b), \(A_2\) lies above the axis, so \(\int_{-1}^{3}f(x)\,dx=A_2=6\). 4. For c), use interval additivity: \(\int_{-4}^{3}f(x)\,dx=-A_1+A_2=-4+6=2\). 5. For d), include all three regions: \(\int_{-4}^{5}f(x)\,dx=-A_1+A_2-A_3=-4+6-2=0\).

a) \(-4\) b) \(6\) c) \(2\) d) \(0\)

- Check what happens when the graph is rotated \(180^\circ\) about the origin. - Compare the areas to the left and right of the \(y\)-axis. - Recall how regions below the \(x\)-axis contribute to a definite integral.

1. The function contains only odd powers of \(x\), so \(f(-x)=-f(x)\). Its graph has point symmetry about the origin. 2. The area between the graph and the \(x\)-axis on \([-3,0]\) equals the area on \([0,3]\). 3. The graph lies above the \(x\)-axis on \((-3,0)\) and below it on \((0,3)\). Therefore, the two equal areas have opposite signs in the definite integral and cancel. 4. Hence, \(\int_{-3}^{3}f(x)\,dx=0\).

The function is odd, so its graph has point symmetry about the origin. The equal areas on \([-3,0]\) and \([0,3]\) contribute with opposite signs, so the net signed area is \(0\).

- Sketch one full period of the sine curve. - What symmetry does a function with only even powers have? - 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.

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.

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.

- Compare the two graphs as horizontal translations. - Shift the function and both bounds by the same amount. - You may also evaluate both integrals directly. - Use nonnegativity to justify uniqueness.

1. The function \((x-4)^4\) is the graph of \((x-2)^4\) shifted \(2\) units to the right. 2. Shifting both integration bounds by \(2\) preserves the integral value: \(\int_1^3(x-2)^4\,dx=\int_3^5(x-4)^4\,dx\). 3. Therefore, \(b=5\). Because the right-hand integrand is nonnegative and positive except at \(x=4\), the accumulated integral is strictly increasing for \(b>3\), so the solution is unique.

\(b=5\)

- Evaluate the sine integral over a complete period. - Separate the sine and constant terms by linearity. - To double one-period accumulation, use an interval spanning two periods. - The starting point may be arbitrary.

1. Over one full period, \(\int_0^{2\pi}\sin(x)\,dx=0\), so \(\int_0^{2\pi}(\sin(x)+2)\,dx=4\pi\). 2. The right-hand side is therefore \(8\pi\). 3. Any interval of length \(4\pi\) contains two full periods of the integrand. The sine contribution is \(0\), and the constant contribution is \(2(4\pi)=8\pi\). 4. Hence, any pair satisfying \(b=a+4\pi\) works. One choice is \(a=0\), \(b=4\pi\).

For example, \(a=0\) and \(b=4\pi\). More generally, any \(b=a+4\pi\) works.

- 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. - Use a rough sketch when it helps you reason about signed area.

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 \(\int_{0}^{4}(x-4)\,dx\) is negative. Reversing the limits changes the sign, so \(\int_{4}^{0}(x-4)\,dx\) is positive.

a) The integral is zero. b) The integral is negative. c) The integral is positive.

- 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 \(a<b\).

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 \(a=0\), \(b=2\sqrt{2}\).

One pair is \((-2, 2)\), justified by odd symmetry. A second pair is \((0, 2\sqrt{2})\), found by solving the antiderivative equation.

- Sketch each graph over the given interval. - Locate the x-intercepts and any relevant symmetry. - Identify which regions contribute positive or negative signed area. - Look for a vertical line that divides a region into mirror-image halves.

1. For a), the line \(y=x-2\) crosses the x-axis at \(x=2\). On \([0, 2]\), it forms a triangle below the x-axis with signed area \(-2\). On \([2, 4]\), it forms a congruent triangle above the x-axis with signed area \(2\). The net signed area is \(0\). 2. For b), the graph of \(y=(x-3)^3\) has point symmetry about \((3, 0)\), and \([1, 5]\) is symmetric about \(x=3\). The negative signed area on \([1, 3]\) and the positive signed area on \([3, 5]\) have equal magnitude, so they cancel. 3. For c), the sine curve on \([0, \pi]\) is symmetric about \(x=\frac{\pi}{2}\). Therefore, the areas on \([0, \frac{\pi}{2}]\) and \([\frac{\pi}{2}, \pi]\) are equal, so the full integral is twice the first half.

a) The two congruent triangular regions have opposite signs, so their signed areas sum to \(0\). b) The graph and interval are symmetric about \((3, 0)\) and \(x=3\), respectively, so the signed areas cancel. c) The sine curve is symmetric about \(x=\frac{\pi}{2}\), so the two half-intervals contribute equal areas.

- Use a property of definite integrals to split the integrand into two simpler parts. - Identify the geometric figure represented by \(y=\sqrt{r^2-x^2}\) over its full domain. - Interpret the integral of a constant function as an area. - Recall the area formula for a circle.

1. Split the integral using the sum rule: \(\int_{-5}^{5}3\,dx+\int_{-5}^{5}\sqrt{25-x^2}\,dx\). 2. The first integral is the area of a rectangle with width \(5-(-5)=10\) and height \(3\), so its area is \(10\cdot 3=30\). 3. The graph of \(y=\sqrt{25-x^2}\) is the upper semicircle of radius \(5\). Its area is \(\frac{1}{2}\pi(5)^2=\frac{25\pi}{2}\). 4. Adding the two areas gives \(30+\frac{25\pi}{2}\).

\(30+\frac{25\pi}{2}\)

- Use the difference rule for definite integrals. - Consider the two functions in the integrand separately. - Identify the geometric regions formed by their graphs on \([0,4]\). - Recall the area formulas for circles and triangles.

1. Use the difference rule to write \(\int_{0}^{4}\sqrt{16-x^2}\,dx-\int_{0}^{4}x\,dx\). 2. The first integral is the area of a quarter circle of radius \(4\), so its value is \(\frac{1}{4}\pi(4)^2=4\pi\). 3. The second integral is the area of a right triangle with base \(4\) and height \(4\), so its value is \(\frac{1}{2}\cdot 4\cdot 4=8\). 4. Subtracting the areas gives \(4\pi-8\).

\(4\pi-8\)

- 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.

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

a) \(16\) b) \(2\) c) \(\frac{3}{2}\)

- 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.

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: \(\int_0^1[(3x+1)^2-(9x^2+1)]\,dx=\int_0^1 6x\,dx\). 3. Evaluate: \([3x^2]_0^1=3\). 4. For c), use additivity over adjacent intervals: \(\int_1^{2.5}\frac{1}{x^2}\,dx+\int_{2.5}^{5}\frac{1}{x^2}\,dx=\int_1^5x^{-2}\,dx\). 5. Evaluate: \([-\frac{1}{x}]_1^5=-\frac{1}{5}+1=\frac{4}{5}\).

a) \(0\) b) \(3\) c) \(\frac{4}{5}\)

- 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 polynomial multiple of \(f\).

1. An antiderivative is \(F(x)=\frac{1}{4}x^4-2x^2\). Since \(F(2)=-4\) and \(F(-2)=-4\), \(\int_{-2}^{2}(x^3-4x)\,dx=F(2)-F(-2)=0\). 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 \(0\).

1. \(\int_{-2}^{2}(x^3-4x)\,dx=0\) 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 \(f\).

- Identify the vertex and axis of symmetry of the parabola. - Compare the two half-intervals across the symmetry axis. - For an algebraic check, find an antiderivative and evaluate both sides. - Use additivity to split the larger interval.

1. Symmetry approach: The graph is a parabola with vertex \((2, -1)\) and axis of symmetry \(x=2\). The intervals \([1, 2]\) and \([2, 3]\) are mirror images across this axis, so \(\int_1^2f(x)\,dx=\int_2^3f(x)\,dx\). By additivity, \(\int_1^3f(x)\,dx=2\int_2^3f(x)\,dx\). 2. Fundamental Theorem approach: One antiderivative is \(F(x)=\frac{1}{3}(x-2)^3-x\). Then \(\int_1^3f(x)\,dx=F(3)-F(1)=-\frac{4}{3}\), while \(2\int_2^3f(x)\,dx=2[F(3)-F(2)]=2(-\frac{2}{3})=-\frac{4}{3}\).

The equation is true because the graph is symmetric about \(x=2\), making the integrals over \([1, 2]\) and \([2, 3]\) equal. Equivalently, both sides evaluate to \(-\frac{4}{3}\).

- 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?

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 \(\int_0^2f(x)\,dx=0\). 3. By additivity, \(\int_0^4f(x)\,dx=\int_0^2f(x)\,dx+\int_2^4f(x)\,dx=\int_2^4f(x)\,dx\). 4. The statement is true.

True. Point symmetry about \((1, 0)\) makes \(\int_0^2f(x)\,dx=0\), so additivity gives \(\int_0^4f(x)\,dx=\int_2^4f(x)\,dx\).

- 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.

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, \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(x)\,dx=\sin\left(\frac{\pi}{2}\right)-\sin\left(-\frac{\pi}{2}\right)=2\). 2. For b), \(\int_{a}^{b}\cos(x)\,dx=\sin(b)-\sin(a)\). 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 \(2\pi\).

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 \(\int_{\pi}^{2\pi}\cos(x)\,dx=0\).

- 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.

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 \(\int_{-1}^{1}(x^2-\frac{1}{3})\,dx=[\frac{1}{3}x^3-\frac{1}{3}x]_{-1}^{1}=0\). 4. However, \(f(-x)=f(x)\), so the function is even rather than odd. This is a counterexample to Liam's claim.

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.

- 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.

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 \(\int_{-1}^{1}g(x)\,dx+\int_{1}^{2}g(x)\,dx\). 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, \(J=G(2)-G(-1)=\left(4-2\right)-\left(\frac{1}{4}-\frac{1}{2}\right)=2+\frac{1}{4}=\frac{9}{4}=2.25\).

1) \(I=0\) because an odd function has canceling signed-area contributions on an interval symmetric about zero. 2) \(J\) is positive. 3) \(J=\frac{9}{4}=2.25\).

- 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.

1. The first integrand is odd, so its integral over \([-4, 4]\) is \(0\). 2. The second integrand is even. Therefore, \(\int_{-1}^{1}(15x^4-6x^2)\,dx=2\int_0^1(15x^4-6x^2)\,dx\). 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 \(2\int_0^3(x^2+1)\,dx=2[\frac{1}{3}x^3+x]_0^3=24\).

1. \(0\) 2. \(2\) 3. \(24\)

- 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.

1. For 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 b), use linearity: \(I=\int_{-3}^{3}f(x)\,dx+\int_{-3}^{3}c\,dx=0+6c=6c\). 3. For 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 \(\int_{-1}^{1}-1\,dx=-2\).

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 \(\int_{-1}^{1}g(x)\,dx=-2\).

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

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

\(4\)

- 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.

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

a) \(8\) b) \(-2\) c) \(2\) d) \(17\)

- 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.

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: \(I=2\int_0^k-6x^2\,dx=2[-2x^3]_0^k=-4k^3\).

a) The odd terms \(x^3\) and \(2x\) integrate to \(0\) over \([-k, k]\), leaving only \(-6x^2\). b) \(I=-4k^3\)

- Separate the even and odd terms. - The odd part contributes \(0\) over the symmetric interval. - Double the integral of the even part over \([0, 2]\). - Apply the Fundamental Theorem to the remaining polynomial.

1. Separate the odd part \(2x^5-4x^3\) from the even part \(3x^2-1\). 2. The odd part integrates to \(0\) over \([-2, 2]\). 3. For the even part, \(\int_{-2}^{2}(3x^2-1)\,dx=2\int_0^2(3x^2-1)\,dx\). 4. An antiderivative is \(x^3-x\), so the value is \(2[x^3-x]_0^2=2(8-2)=12\).

\(12\)

- Separate the odd and even parts of the integrand. - The odd part contributes \(0\) over a symmetric interval. - Double the integral of the even part over \([0, 2]\).

1. Separate the odd term \(\frac{1}{2}x^3\) from the even part \(-x^2+2\). 2. The odd term integrates to \(0\) over \([-2, 2]\). 3. For the even part, \(\int_{-2}^{2}(-x^2+2)\,dx=2\int_0^2(-x^2+2)\,dx\). 4. An antiderivative is \(-\frac{1}{3}x^3+2x\), so \(2[-\frac{1}{3}x^3+2x]_0^2=2(-\frac{8}{3}+4)=\frac{8}{3}\).

\(\frac{8}{3}\)

- 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.

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

a) \(\int_{-3}^{2}f(x)\,dx=1.25\) b) \(\int_{-1}^{5}f(x)\,dx=-2.60\) c) \(\int_{5}^{-3}f(x)\,dx=4.10\)

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

1. For a), point symmetry about \(P(2,0)\) makes the region above the \(x\)-axis on \([0,2]\) congruent to the region below the axis on \([2,4]\). Their signed contributions are opposites, so \(\int_{0}^{4}f(x)\,dx=0\). 2. For b), an antiderivative is \(F(x)=\frac{1}{8}x^4-x^3+2x^2\). The positive contribution on \([0,2]\) is \(F(2)-F(0)=2\). By symmetry, the region on \([2,4]\) also has area \(2\). Therefore, the total area is \(2+2=4\) square units. 3. For c), use linearity: \(\int_{0}^{4}g(x)\,dx=\int_{0}^{4}f(x)\,dx+\int_{0}^{4}1\,dx=0+4=4\). Geometrically, shifting the function upward by \(1\) adds a rectangle of width \(4\) and height \(1\) to the net signed area.

a) The congruent positive and negative signed-area contributions cancel, so \(\int_{0}^{4}f(x)\,dx=0\). b) The total area is \(4\) square units. c) \(\int_{0}^{4}g(x)\,dx=\int_{0}^{4}f(x)\,dx+\int_{0}^{4}1\,dx=0+4=4\).

- 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.

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.

a) True b) True c) True d) False

- 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.

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

a) \(1.2\) b) \(-2.7\) c) \(-1.5\) d) \(3.5\)

- Compare the two formulas and identify the horizontal shift. - Compare the endpoints of the two intervals. - Decide whether a horizontal translation changes the area of a region.

1. The graph of \(g\) is the graph of \(f\) shifted \(2\) units to the right because \(g(x)=f(x-2)\). 2. The interval \([1,3]\) is also the interval \([-1,1]\) shifted \(2\) units to the right. 3. Shifting both a graph and its interval horizontally does not change the area between the graph and the x-axis. Therefore, the two regions are congruent and the definite integrals are equal.

The graph and the interval are both shifted \(2\) units to the right, so the two regions have the same area. Therefore, the definite integrals are equal.

- 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.

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, \(\int_{-3}^{3}h(x)\,dx=H(3)-H(-3)=6-(-6)=12\). Since the graph lies above the \(x\)-axis on \([-3,3]\), this value is also the geometric area.

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) \(\int_{-3}^{3}h(x)\,dx=12\), which is the area between the graph and the \(x\)-axis.

- 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.

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 \(\int_{0}^{4}f(x)\,dx=-16\). The area is positive, so the corrected statement is \(A_2=-\int_{0}^{4}f(x)\,dx\). 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.

(1) Correct (2) False; \(A_2=-\int_{0}^{4}f(x)\,dx\) (3) Correct (4) Correct

- 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.

1. For \(f(x)=3x^2+5\), \(\int_{-k}^k(3x^2+5)\,dx=[x^3+5x]_{-k}^k=2k^3+10k\). Also, \(2\int_0^k(3x^2+5)\,dx=2[x^3+5x]_0^k=2k^3+10k\). 2. The equation is equivalent to \(\int_{-k}^{0}f(x)\,dx=\int_0^k f(x)\,dx\) 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 \(c=0\).

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. \(c=0\)

- 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.

1. Substitution gives \(f_k(0)=0^4+k\cdot 0^2=0\). 2. Compute the integral: \(\int_{-2}^{2}(x^4+kx^2)\,dx=\frac{64}{5}+\frac{16k}{3}\). 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.

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.

- 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.

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 \([1, 4]\): \(\int_1^4[(3-k)g(x)-(k+2)]\,dx\). 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: \(\int_1^4[-(3+2)]\,dx=\int_1^4-5\,dx=-15\). Therefore, \(k=3\) works.

\(k=3\)

- 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.

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

a) \(12\) b) \(-12\) c) \(12\) d) \(0\) e) Cannot be determined

- 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.

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, \(\int_{-1}^{2}f(x)\,dx=\int_{-1}^{1}f(x)\,dx+\int_{1}^{2}f(x)\,dx\). The first integral is \(0\) because \(f\) is odd and \([-1,1]\) is symmetric about zero. Therefore, \(\int_{-1}^{2}f(x)\,dx=\int_{1}^{2}f(x)\,dx\).

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) \(\int_{-2.5}^{2.5}f(x)\,dx=0\). 4) Since \(\int_{-1}^{1}f(x)\,dx=0\), \(\int_{-1}^{2}f(x)\,dx=\int_{1}^{2}f(x)\,dx\).