Applications of Integration: practice solutions, College – download the PDF

1 Area between a line and a parabola ★★★

Solve \(x=x^2\): \(x=0\) or \(x=1\). On \((0,1)\) the line is above the parabola (try \(x=\tfrac12\): \(0.5>0.25\)).

\(A=\int_0^1(x-x^2)\,dx=\Big[\tfrac{x^2}{2}-\tfrac{x^3}{3}\Big]_0^1=\tfrac12-\tfrac13=\tfrac16\).

The area is \(\tfrac16\) square unit.

2 Area under a roof line ★★★

The curves meet when \(x^2=4\), so \(x=\pm2\). The top curve is \(y=4\).

\(A=\int_{-2}^{2}(4-x^2)\,dx=\Big[4x-\tfrac{x^3}{3}\Big]_{-2}^{2}=\tfrac{16}{3}+\tfrac{16}{3}=\tfrac{32}{3}\).

The window has area \(\tfrac{32}{3}\approx 10.67\) ft\(^2\).

3 A cone from a line ★★★

Each disk has radius \(x\): \(V=\pi\int_0^3 x^2\,dx=\pi\cdot\tfrac{27}{3}=9\pi\).

Check: the solid is a cone with radius 3 and height 3, so \(V=\tfrac13\pi r^2h=\tfrac13\pi(9)(3)=9\pi\). The volume is \(9\pi\approx 28.27\) cubic units.

4 Length of a ramp ★★★

\(f'(x)=3\), so \(L=\int_0^4\sqrt{1+9}\,dx=4\sqrt{10}\approx 12.65\).

Check: the endpoints are \((0,1)\) and \((4,13)\), so the distance is \(\sqrt{4^2+12^2}=\sqrt{160}=4\sqrt{10}\). Both agree.

5 Average of a linear function ★★★

\(\int_0^4(2x+1)\,dx=[x^2+x]_0^4=20\), so \(f_{\text{avg}}=\tfrac{20}{4}=5\).

Solve \(2x+1=5\): \(x=2\), the midpoint, as expected for a straight line.

6 Constant force ★★★

\(W=\int_0^{12}25\,dx=25\cdot12=300\) ft\(\cdot\)lb.

In joules: \(300\times1.3558\approx 406.7\) J.

7 Checking a differential equation ★★★

\(y'=-2Ce^{-2x}=-2y\), so every \(C\) works.

\(y(0)=C=5\), so \(y=5e^{-2x}\) and \(y(1)=5e^{-2}\approx 0.677\).

8 Square root against a line ★★★

\(\sqrt{x}=\tfrac x2\) gives \(x=0\) or \(x=4\). On \((0,4)\) the square root is on top (at \(x=1\): \(1>0.5\)).

\(A=\int_0^4\big(x^{1/2}-\tfrac x2\big)dx=\Big[\tfrac23x^{3/2}-\tfrac{x^2}{4}\Big]_0^4=\tfrac{16}{3}-4=\tfrac43\).

9 Curves that cross ★★★

\(x=x^3\) gives \(x=-1,0,1\). On \((0,1)\) the line is on top; on \((-1,0)\) the cubic is on top, so we must split.

By symmetry \(A=2\int_0^1(x-x^3)\,dx=2\Big(\tfrac12-\tfrac14\Big)=\tfrac12\).

10 Washer about the x-axis ★★★

The curves meet at \(x=0\) and \(x=2\); \(2x\) is the outer radius and \(x^2\) the inner radius.

\(V=\pi\int_0^2(4x^2-x^4)\,dx=\pi\Big(\tfrac{32}{3}-\tfrac{32}{5}\Big)=\tfrac{64\pi}{15}\approx 13.40\).

11 A cubic bowl ★★★

\(V=\pi\int_0^1x^6\,dx=\pi\cdot\tfrac17=\tfrac{\pi}{7}\approx 0.449\) cubic unit.

12 A vase by shells ★★★

Radius \(x\), height \(4x-x^2\): \(V=2\pi\int_0^4(4x^2-x^3)\,dx=2\pi\Big[\tfrac{4x^3}{3}-\tfrac{x^4}{4}\Big]_0^4=2\pi\Big(\tfrac{256}{3}-64\Big)=\tfrac{128\pi}{3}\approx 134.0\).

13 Length of a curved cable ★★★

\(y'=\tfrac32\sqrt{x}\), so \(1+(y')^2=1+\tfrac94x\).

\(L=\int_0^4\sqrt{1+\tfrac94x}\,dx=\tfrac{8}{27}\Big[\big(1+\tfrac94x\big)^{3/2}\Big]_0^4=\tfrac{8}{27}(10^{3/2}-1)\approx 9.07\) ft.

14 Stretching a spring in metric units ★★★

\(W=\int_{0.1}^{0.3}200x\,dx=100x^2\Big|_{0.1}^{0.3}=100(0.09-0.01)=8\) J.

15 Average temperature ★★★

\(\int_0^{12}\sin\!\big(\tfrac{\pi t}{12}\big)dt=\tfrac{12}{\pi}\big[-\cos\tfrac{\pi t}{12}\big]_0^{12}=\tfrac{24}{\pi}\).

\(T_{\text{avg}}=60+\tfrac{20}{12}\cdot\tfrac{24}{\pi}=60+\tfrac{40}{\pi}\approx 72.7^\circ\)F (about 22.6 \(^\circ\)C).

16 Integrating with respect to y ★★★

Solve \(y^2=y+6\): \((y-3)(y+2)=0\), so \(y=-2\) or \(y=3\). The line is to the right of the parabola in between.

\(A=\int_{-2}^{3}(y+6-y^2)\,dy=\Big[\tfrac{y^2}{2}+6y-\tfrac{y^3}{3}\Big]_{-2}^{3}=\tfrac{27}{2}+\tfrac{22}{3}=\tfrac{125}{6}\approx 20.83\).

17 Shells about a vertical line ★★★

A strip at \(x\) is at distance \(2-x\) from the axis and has height \(x^2\).

\(V=2\pi\int_0^1(2-x)x^2\,dx=2\pi\Big(\tfrac23-\tfrac14\Big)=\tfrac{5\pi}{6}\approx 2.62\).

18 Washers about a horizontal line ★★★

The distances to the axis are \(R=x+1\) and \(r=x^2+1\).

\(V=\pi\int_0^1\big[(x+1)^2-(x^2+1)^2\big]dx=\pi\int_0^1(2x-x^2-x^4)\,dx=\pi\Big(1-\tfrac13-\tfrac15\Big)=\tfrac{7\pi}{15}\).

19 An arc that needs a table ★★★

\(y'=x\), so \(L=\int_0^1\sqrt{1+x^2}\,dx=\tfrac12\big(\sqrt2+\ln(1+\sqrt2)\big)\approx 1.148\).

20 Pumping water from a tank ★★★

A layer at height \(y\) above the bottom has volume \(9\pi\,dy\), weighs \(62.4\cdot9\pi\,dy\) pounds and is lifted \(8-y\) feet.

\(W=\int_0^8 62.4\cdot9\pi\,(8-y)\,dy=561.6\pi\cdot32=17{,}971.2\pi\approx 56{,}458\) ft\(\cdot\)lb.

21 A separable initial-value problem ★★★

Separate: \(e^y\,dy=x\,dx\). Integrate: \(e^y=\tfrac{x^2}{2}+C\). From \(y(0)=0\): \(1=C\).

So \(y=\ln\!\big(1+\tfrac{x^2}{2}\big)\) and \(y(2)=\ln3\approx 1.099\).

22 Cooling coffee ★★★

Separate: \(\dfrac{dT}{T-70}=-k\,dt\), so \(T-70=Ae^{-kt}\) with \(A=120\).

At \(t=10\): \(80=120e^{-10k}\), so \(e^{-10k}=\tfrac23\). At \(t=20\): \(T=70+120\cdot\tfrac49=\tfrac{370}{3}\approx 123.3^\circ\)F (about 50.7 \(^\circ\)C).

For \(T=100\): \(30=120(2/3)^{t/10}\), so \(t=10\,\dfrac{\ln4}{\ln1.5}\approx 34.2\) minutes.