Free math worksheets

- The y-axis is the left boundary \(x=0\). - The horizontal width is \(\sqrt y\). - The given horizontal lines provide the integration limits.

1. The horizontal width is \(\sqrt y-0\). 2. Therefore, \(A=\int_0^4\sqrt y\,\mathrm{d}y=\frac{16}{3}\).

\(\frac{16}{3}\) square units

- Relate average horizontal width to the integral of width. - Use the full vertical length.

1. The vertical extent is \(5-(-2)=7\) units. 2. Average width equals area divided by vertical extent. 3. Therefore, \(A=9\cdot7=63\) square units.

\(63\) square units

- For a horizontal slice, subtract the left x-value from the right x-value. - Simplify the width before integrating. - The y-limits are already given.

1. The right boundary is \(x=1+\sin y\), and the left boundary is \(x=\sin y-2\). 2. Their horizontal separation is \((1+\sin y)-(\sin y-2)=3\). 3. Therefore, \(A=\int_{-\pi/2}^{\pi/2}3\,\mathrm{d}y=3\pi\).

\(3\pi\) square units

- Use the horizontal boundaries \(y=0\) and \(y=2\) as the limits. - The horizontal width is the right x-value minus the left x-value. - Integrate \(e^y-1\) with respect to \(y\).

1. For \(0\le y\le2\), \(x=e^y\) is the right boundary and \(x=1\) is the left boundary. 2. Therefore, \(A=\int_0^2(e^y-1)\,\mathrm{d}y\). 3. Evaluating gives \(A=e^2-3\).

\(e^2-3\) square units

- Solve \(y^3=y\) to find all intersection heights. - The two regions correspond under a \(180^\circ\) rotation about the origin. - Integrate right boundary minus left boundary for \(0\le y\le1\), then double.

1. The curves meet when \(y^3=y\), so \(y=-1\), \(y=0\), and \(y=1\). 2. The two regions have equal areas because the graphs are symmetric about the origin. 3. For \(0\le y\le1\), \(x=y\) is the right boundary and \(x=y^3\) is the left boundary. 4. Therefore, \(A=2\int_0^1(y-y^3)\,\mathrm{d}y=\frac{1}{2}\).

\(\frac{1}{2}\) square units

- The y-axis has equation \(x=0\). - The horizontal width is \(\cos y\). - Use the given y-values as the integration limits.

1. The right boundary is \(x=\cos y\), and the left boundary is \(x=0\). 2. Therefore, \(A=\int_{-\pi/2}^{\pi/2}\cos y\,\mathrm{d}y=2\).

\(2\) square units

- The right-boundary rule changes at \(y=2\). - Each horizontal width is the stated x-value minus \(0\). - Add the two integrals rather than combining the formulas across the wrong interval.

1. Use the stated right boundary on each y-interval. 2. Therefore, \(A=\int_0^2 2y\,\mathrm{d}y+\int_2^5(6-y)\,\mathrm{d}y\). 3. The two contributions are \(4\) and \(\frac{15}{2}\), so \(A=\frac{23}{2}\).

\(\frac{23}{2}\) square units

- Choose the positive x-branch of the circle. - A horizontal slice runs from \(x=0\) to \(x=\sqrt{25-y^2}\). - Recognize the resulting integral as the area of a semicircle.

1. The right semicircle is \(x=\sqrt{25-y^2}\), with \(-5\le y\le5\). The y-axis is the left boundary. 2. Therefore, \(A=\int_{-5}^{5}\sqrt{25-y^2}\,\mathrm{d}y\). 3. This integral represents a semicircle of radius \(5\), so \(A=\frac{25\pi}{2}\).

\(\frac{25\pi}{2}\) square units

- Treat horizontal width as the function being integrated. - Use the actual change in \(y\) for each trapezoid width. - Add the four trapezoidal estimates.

1. Apply the trapezoidal rule to the width function using the unequal y-intervals. 2. The estimate is \(\frac{1}{2}(2+5)+\frac{2}{2}(5+4)+\frac{1}{2}(4+6)+\frac{3}{2}(6+1)=28\). 3. Therefore, the area is approximately \(28\) square units.

\(\approx28\) square units

- Treat \(w(y)\) as the horizontal width of a thin strip. - Integrate the width over the full north-south interval. - Square meters result from multiplying meters of width by meters of north-south distance.

1. The width function is already the right-minus-left distance at each north-south coordinate. 2. Therefore, \(A=\int_0^8[120-5(y-4)^2]\,\mathrm{d}y=\frac{2240}{3}\,\text{m}^2\).

\(\frac{2240}{3}\,\text{m}^2\)

- At each height, subtract the left x-coordinate from the right x-coordinate. - Use \(y=0\) and \(y=4\) as the limits. - Integrating feet of width over feet of height gives square feet.

1. At height \(y\), the horizontal width is \(\left(6-\frac{y}{2}\right)-\frac{y^2}{8}\). 2. Therefore, \(A=\int_0^4\left(6-\frac{y}{2}-\frac{y^2}{8}\right)\,\mathrm{d}y\). 3. Evaluating gives \(A=24-4-\frac{8}{3}=\frac{52}{3}\,\text{ft}^2\).

\(\frac{52}{3}\,\text{ft}^2\)

- Compare the horizontal widths rather than the absolute x-locations. - A common horizontal translation affects both boundaries equally.

1. Region A has width \(f(y)-g(y)\). 2. Region B has width \([f(y)+7]-[g(y)+7]=f(y)-g(y)\). 3. The width functions and limits are identical, so the areas are equal.

The two regions have equal area.

- Write the exact definite integral before using a calculator. - The y-axis is the left boundary \(x=0\). - Use numerical integration because this integrand has no elementary antiderivative.

1. The horizontal width is \(e^{-y^2}\). 2. Therefore, \(A=\int_0^1e^{-y^2}\,\mathrm{d}y\). 3. Numerical evaluation gives \(A\approx0.747\).

\(\approx0.747\) square units

- Rewrite each boundary as x in terms of y. - A horizontal slice runs from \(x=\frac{y}{2}\) to \(x=\sqrt y\). - Use the y-coordinates of the intersections as the limits.

1. Rewrite the first-quadrant branches as \(x=\sqrt y\) and \(x=\frac{y}{2}\). 2. The curves meet at \(y=0\) and \(y=4\). For \(0<y<4\), \(x=\sqrt y\) is the right boundary. 3. Therefore, \(A=\int_0^4\left(\sqrt y-\frac{y}{2}\right)\,\mathrm{d}y=\frac{4}{3}\).

\(\frac{4}{3}\) square units

- Use symmetry across the x-axis. - On the upper half of the region, \(|y|=y\). - For each horizontal slice, subtract the left x-value from the right x-value.

1. For \(y\ge0\), the curves intersect when \(y=2-y^2\), which gives \(y=1\). By symmetry, the other intersection occurs at \(y=-1\). 2. The right boundary is \(x=2-y^2\), and the left boundary is \(x=|y|\). 3. By symmetry, \(A=2\int_0^1(2-y^2-y)\,\mathrm{d}y=\frac{7}{3}\).

\(\frac{7}{3}\) square units

- Express the intersection heights in terms of \(k\). - A horizontal slice has width \(k-y^2\). - Use symmetry before solving the area equation.

1. The intersections occur at \(y=\pm\sqrt k\). 2. The area is \(A(k)=\int_{-\sqrt k}^{\sqrt k}(k-y^2)\,\mathrm{d}y=\frac{4}{3}k^{3/2}\). 3. Set \(\frac{4}{3}k^{3/2}=36\). Then \(k^{3/2}=27\), so \(k=9\).

\(k=9\)

- A horizontal slice has width \(y-y^2\). - Use \(0\) and \(c\) as the integration limits. - After solving the resulting cubic equation, apply the condition \(0<c<1\).

1. On \([0,c]\), \(x=y\) is the right boundary and \(x=y^2\) is the left boundary. 2. The area equation is \(\int_0^c(y-y^2)\,\mathrm{d}y=\frac{c^2}{2}-\frac{c^3}{3}=\frac{1}{12}\). 3. Rearranging gives \(4c^3-6c^2+1=0=(2c-1)(2c^2-2c-1)\). 4. The roots are \(\frac{1}{2}\) and \(\frac{1\pm\sqrt3}{2}\). Only \(c=\frac{1}{2}\) lies in \((0,1)\).

\(c=\frac{1}{2}\)

- Compare how the two curves are already written. - Set the x-expressions equal to find the y-limits. - For each horizontal slice, subtract the left boundary from the right boundary.

1. Horizontal slices use one right-minus-left expression. Vertical slices would require solving for upper and lower branches and splitting the region. 2. The intersections satisfy \(y^2-2=2-y^2\), so \(y=\pm\sqrt2\). 3. Therefore, \(A=\int_{-\sqrt2}^{\sqrt2}\left[(2-y^2)-(y^2-2)\right]\,\mathrm{d}y=\frac{16\sqrt2}{3}\).

a) Horizontal slices use one integral with the given x-boundaries. b) \(\frac{16\sqrt2}{3}\) square units

- Find the height where the two x-boundaries are equal. - Test which expression is farther right below and above that height. - Geometric area uses a nonnegative width on each interval.

1. The boundaries switch order where \(y+1=5-y\), so \(y=2\). 2. The student's width is negative below \(y=2\) and positive above \(y=2\), so the integral allows cancellation instead of adding geometric areas. 3. The correct area is \(A=\int_0^2[(5-y)-(y+1)]\,\mathrm{d}y+\int_2^3[(y+1)-(5-y)]\,\mathrm{d}y=5\).

The right and left boundaries switch at \(y=2\); the correct area is \(5\) square units.

- Solve the logarithmic boundary for x. - The fixed line \(x=e^2\) is the right boundary. - Use \(y=0\) and \(y=2\) as the integration limits.

1. Rewrite \(y=\ln x\) as \(x=e^y\). 2. For \(0\le y\le2\), a horizontal slice runs from \(x=e^y\) to \(x=e^2\). 3. Therefore, \(A=\int_0^2(e^2-e^y)\,\mathrm{d}y=e^2+1\).

\(e^2+1\) square units

- Set the two x-expressions equal to find the y-limits. - Test which x-expression is larger between the intersections. - Integrate right boundary minus left boundary.

1. Solve \(y^2=2y+3\): \((y-3)(y+1)=0\). The intersection heights are \(y=-1\) and \(y=3\). 2. For \(-1\le y\le3\), \(x=2y+3\) is the right boundary and \(x=y^2\) is the left boundary. 3. Therefore, \(A=\int_{-1}^{3}(2y+3-y^2)\,\mathrm{d}y=\frac{32}{3}\).

a) \(y=-1\) and \(y=3\) b) \(\frac{32}{3}\) square units

- First calculate half of the total area. - Use the unknown height \(c\) as the upper limit of an accumulation integral. - Solve the resulting equation numerically and retain the solution in \((0,2)\).

1. The total area is \(\int_0^2(4-y^2)\,\mathrm{d}y=\frac{16}{3}\). 2. The lower portion must have area \(\frac{8}{3}\), so \(\int_0^c(4-y^2)\,\mathrm{d}y=\frac{8}{3}\). 3. This gives \(4c-\frac{c^3}{3}=\frac{8}{3}\). 4. The solution in \((0,2)\) is \(c\approx0.695\).

\(c\approx0.695\)

- Solve the ellipse equation for both x-boundaries. - Subtract the negative branch from the positive branch to get the width. - A substitution \(u=\frac{y}{3}\) converts the integral to a semicircle-area integral.

1. Solve for the horizontal boundaries: \(x=\pm4\sqrt{1-\frac{y^2}{9}}\). 2. The horizontal width is \(8\sqrt{1-\frac{y^2}{9}}\) for \(-3\le y\le3\). 3. Therefore, \(A=\int_{-3}^{3}8\sqrt{1-\frac{y^2}{9}}\,\mathrm{d}y=12\pi\).

\(12\pi\) square units

- Find the height where the two x-boundaries are equal. - The signed integral preserves the order \(\frac{1}{y}-y\) across the whole interval. - For total area, use right boundary minus left boundary separately below and above \(y=1\).

1. The curves switch right-left order at \(y=1\). 2. The signed integral is \(\left[\ln y-\frac{y^2}{2}\right]_{1/2}^{2}=\ln4-\frac{15}{8}\). 3. The total area is \(\int_{1/2}^{1}\left(\frac{1}{y}-y\right)\,\mathrm{d}y+\int_1^2\left(y-\frac{1}{y}\right)\,\mathrm{d}y=\frac{9}{8}\). 4. The signed integral allows cancellation after the boundaries switch, while total area uses a nonnegative width on both subintervals.

Signed integral: \(\ln4-\frac{15}{8}\) Total area: \(\frac{9}{8}\) square units

- Write one trapezoidal-area expression for each y-interval. - Use interval lengths \(2\) and \(3\), not a common step size. - Set the sum equal to the stated estimate and solve the linear equation.

1. The two trapezoids give \(\frac{2}{2}(3+m)+\frac{3}{2}(m+7)=\frac{27}{2}+\frac{5}{2}m\). 2. Set this estimate equal to \(25\). 3. Solving gives \(m=\frac{23}{5}\).

\(m=\frac{23}{5}\)

- Find both intersection heights in terms of \(a\). - A horizontal slice has width \(ay-y^2\). - Set the parameterized area equal to \(\frac{9}{2}\) and use \(a>0\).

1. The intersections satisfy \(ay=y^2\), so \(y=0\) and \(y=a\). 2. On \(0\le y\le a\), the line is the right boundary. Thus, \(A(a)=\int_0^a(ay-y^2)\,\mathrm{d}y=\frac{a^3}{6}\). 3. Set \(\frac{a^3}{6}=\frac{9}{2}\). Then \(a^3=27\), so \(a=3\).

\(a=3\)

- Use only the branch of \(y=x^2\) that lies in the first quadrant. - Solve the two equations together to find the y-limits. - For a test height between \(0\) and \(1\), compare \(\sqrt y\) and \(y^2\).

1. Rewrite \(y=x^2\) as \(x=\sqrt y\) in the first quadrant. 2. The curves intersect at \(y=0\) and \(y=1\). 3. For \(0\le y\le1\), \(x=\sqrt y\) is the right boundary and \(x=y^2\) is the left boundary. 4. Therefore, \(A=\int_0^1(\sqrt y-y^2)\,\mathrm{d}y=\frac{1}{3}\).

\(\frac{1}{3}\) square units

- Center the symmetry at \(y=1\). - Set the two x-expressions equal to find the y-limits. - Subtract the left boundary from the right boundary.

1. The intersections satisfy \(2(y-1)^2=3\), so \(y=1\pm\sqrt{\frac{3}{2}}\). 2. The right boundary is \(x=3-(y-1)^2\), and the left boundary is \(x=(y-1)^2\). 3. Therefore, \(A=\int_{1-\sqrt{3/2}}^{1+\sqrt{3/2}}[3-2(y-1)^2]\,\mathrm{d}y=2\sqrt6\).

\(2\sqrt6\) square units

- The y-axis is the left boundary \(x=0\). - Use integration by parts to find an antiderivative of \(\ln y\). - Evaluate the antiderivative at \(y=e^2\) and \(y=1\).

1. The horizontal width is \(\ln y-0=\ln y\). 2. Therefore, \(A=\int_1^{e^2}\ln y\,\mathrm{d}y\). 3. Using the antiderivative \(y\ln y-y\), the area is \(e^2+1\).

\(e^2+1\) square units

- Set the two candidate right boundaries equal to find the switch height. - Test which candidate is farther right on each side of the switch. - Subtract \(y^2\), the left boundary, in both integrals.

1. The candidate right boundaries meet when \(4y=y+6\), so \(y=2\). 2. For \(0\le y\le2\), \(x=y+6\) is farther right. For \(2\le y\le3\), \(x=4y\) is farther right. 3. Therefore, \(A=\int_0^2(y+6-y^2)\,\mathrm{d}y+\int_2^3(4y-y^2)\,\mathrm{d}y=15\).

\(15\) square units

- Write each slanted edge as x in terms of y. - Subtract the left-edge formula from the right-edge formula. - Use \(0\le y\le4\) as the integration interval.

1. The left edge from \((0,0)\) to \((2,4)\) has equation \(x=\frac{y}{2}\). 2. The right edge from \((6,0)\) to \((4,4)\) has equation \(x=6-\frac{y}{2}\). 3. The horizontal width is \(6-y\) for \(0\le y\le4\). 4. Therefore, \(A=\int_0^4(6-y)\,\mathrm{d}y=16\).

\(16\) square units

- A linear width function makes the region a trapezoid when viewed horizontally. - Use the area and vertical extent to recover the missing endpoint width. - Width equals the right-boundary x-value minus the left-boundary x-value.

1. A linear width has area equal to the vertical extent times the mean of its endpoint widths. 2. Thus, \(30=6\cdot\frac{3+w(7)}{2}\), giving \(w(7)=7\). 3. The linear width through \((1,3)\) and \((7,7)\) is \(w(y)=\frac{2}{3}y+\frac{7}{3}\). 4. Since \(w(y)=R(y)-(-1)\), the right boundary is \(R(y)=\frac{2}{3}y+\frac{4}{3}\).

a) \(w(7)=7\) b) \(x=\frac{2}{3}y+\frac{4}{3}\)

- Use \(w'(y)\) and \(w(0)\) to recover the width function. - The area is the accumulation of horizontal width over the y-interval. - Keep the two integrations conceptually separate: first find width, then find area.

1. Accumulate the width change: \(w(y)=3+\int_0^y(2u+1)\,\mathrm{d}u=y^2+y+3\). 2. The region's area is \(\int_0^2w(y)\,\mathrm{d}y\). 3. Therefore, \(A=\int_0^2(y^2+y+3)\,\mathrm{d}y=\frac{32}{3}\).

\(\frac{32}{3}\) square units

- Write x as a function of y for each slanted side. - Subtract the left-side formula from the right-side formula. - Use the top vertex height as the upper integration limit.

1. The left side has equation \(x=\frac{3}{5}y\). 2. The right side has equation \(x=8-y\). 3. The horizontal width is \(8-\frac{8}{5}y\) for \(0\le y\le5\). 4. Therefore, \(A=\int_0^5\left(8-\frac{8}{5}y\right)\,\mathrm{d}y=20\).

\(20\) square units

- First calculate the area of the original region. - A horizontal scale factor and a vertical scale factor multiply to give the area scale factor. - Apply the scale factor to the original area rather than reconstructing the transformed integral.

1. The area of \(R\) is \(\int_{-2}^{2}(4-y^2)\,\mathrm{d}y=\frac{32}{3}\). 2. Multiplying x-coordinates by \(3\) multiplies area by \(3\). Multiplying y-coordinates by \(\frac{1}{2}\) multiplies area by \(\frac{1}{2}\). 3. The combined area scale factor is \(\frac{3}{2}\), so \(A_S=\frac{3}{2}\cdot\frac{32}{3}=16\).

\(16\) square units

- Compare \(\tan y\) and \(y\) on the stated interval. - Use \(-\ln|\cos y|\) as an antiderivative of \(\tan y\). - Keep the logarithmic and \(\pi\)-terms exact.

1. On the interval, \(\tan y\ge y\), so the horizontal width is \(\tan y-y\). 2. Therefore, \(A=\int_0^{\pi/4}(\tan y-y)\,\mathrm{d}y\). 3. Evaluating gives \(A=\frac{\ln2}{2}-\frac{\pi^2}{32}\).

\(\frac{\ln2}{2}-\frac{\pi^2}{32}\) square units

- The horizontal width is the distance from the y-axis to \(x=\arcsin y\). - Use integration by parts, or verify the supplied antiderivative by differentiating. - Evaluate the inverse-sine values at \(0\) and \(1\).

1. For \(0\le y\le1\), the right boundary is \(x=\arcsin y\), and the left boundary is \(x=0\). 2. Therefore, \(A=\int_0^1\arcsin y\,\mathrm{d}y\). 3. An antiderivative is \(y\arcsin y+\sqrt{1-y^2}\). 4. Evaluating from \(0\) to \(1\) gives \(A=\frac{\pi}{2}-1\).

\(\frac{\pi}{2}-1\) square units

- Identify where each curve meets the y-axis. - Find the height where the two curved boundaries meet. - The y-axis is the left boundary in both integrals.

1. The parabola meets the y-axis at \(y=0\), the line meets it at \(y=2\), and the parabola meets the line at \(y=1\). 2. For \(0\le y\le1\), the right boundary is \(x=y^2\). 3. For \(1\le y\le2\), the right boundary is \(x=2-y\). 4. One valid setup is \(\int_0^1y^2\,\mathrm{d}y+\int_1^2(2-y)\,\mathrm{d}y\).

\(\int_0^1y^2\,\mathrm{d}y+\int_1^2(2-y)\,\mathrm{d}y\)

- Verify the two intersection heights. - At a sample height between \(-3\) and \(2\), compare the two x-values. - A positive area setup uses increasing limits and right boundary minus left boundary.

1. The intersections satisfy \(y^2=6-y\), giving \(y=-3\) and \(y=2\). 2. Between them, \(x=6-y\) lies to the right of \(x=y^2\). Choice B uses right minus left with increasing limits. 3. The area is \(\int_{-3}^{2}(6-y-y^2)\,\mathrm{d}y=\frac{125}{6}\).

Choice B; the area is \(\frac{125}{6}\) square units.

- Write one horizontal-width integral for each half of the region. - Translate “twice the area” into an equation with the upper and lower areas. - Check that the resulting line stays on or to the right of the y-axis.

1. The lower area is \(\int_{-2}^{0}(4+my)\,\mathrm{d}y=8-2m\). 2. The upper area is \(\int_0^{2}(4+my)\,\mathrm{d}y=8+2m\). 3. Require \(8+2m=2(8-2m)\). Then \(6m=8\), so \(m=\frac43\).

\(m=\frac43\)

- Solve each curve for x in terms of y. - Below \(y=1\), both sides of the parabola bound the slice. - Above \(y=1\), the line becomes the right boundary.

1. The intersections are \((-2,4)\) and \((1,1)\). Horizontal slices change their right boundary at \(y=1\). 2. For \(0\le y\le1\), a slice runs from \(x=-\sqrt y\) to \(x=\sqrt y\). 3. For \(1\le y\le4\), a slice runs from \(x=-\sqrt y\) to \(x=2-y\). 4. Therefore, \(A=\int_0^1 2\sqrt y\,\mathrm{d}y+\int_1^4(2-y+\sqrt y)\,\mathrm{d}y=\frac{9}{2}\).

\(\frac{9}{2}\) square units