Further integration: practice Problem set B Worked solutions

\(\frac{dy}{dx}=21\sqrt{x}\), and y = 400 when x = 9. Find y in terms of x.

• \(y=\int 21x^{1/2}\,dx=14x^{3/2}+c\)
• \(x=9, y=400: 400=14(9)^{3/2}+c=378+c\)
• \(c=22\)
• \(y=14x^{3/2}+22\)

Find the required function in each case.

1. (a) \(f'(x)=4\cos x-7\sin x\) and f(0) = 10. Find f(x).

   • \(f(x)=4\sin x+7\cos x+c\) (the integral of \(-\sin x\) is \(\cos x\))
   • \(f(0)=7+c=10\)
   • \(c=3\)
   • \(f(x)=4\sin x+7\cos x+3\)

2. (b) \(\frac{dy}{dx}=5e^x-\frac{2}{x}\) for x > 0, and y = 1 when x = 1. Find y in terms of x.

   • \(y=5e^x-2\ln x+c\)
   • \(x=1: 1=5e-2\ln 1+c=5e+c\)
   • \(c=1-5e\)
   • \(y=5e^x-2\ln x+1-5e\), which is \(y=5e^x-2\ln x-12.6\) (3 s.f.)

3. (c) \(\int\frac{8x^2+6}{2x}\,dx\)

   • \(\frac{8x^2+6}{2x}=4x+\frac{3}{x}\) (divide each term by the denominator)
   • \(\int\left(4x+\frac{3}{x}\right)dx=2x^2+3\ln|x|+c\)

Find each integral.

1. (a) \(\int(2-5x)^{7}\,dx\)

   • \(u=2-5x,\ \frac{du}{dx}=-5\), so \(dx=-\frac{1}{5}du\)
   • \(\int u^{7}\left(-\frac{1}{5}\right)du=-\frac{1}{5}\cdot\frac{u^{8}}{8}\)
   • \(=-\frac{1}{40}(2-5x)^{8}+c\)

2. (b) \(\int(5\sin(3x)+4\cos(6x))\,dx\)

   • \(\int 5\sin(3x)\,dx=-\frac{5}{3}\cos(3x)\) (divide by the coefficient of x)
   • \(\int 4\cos(6x)\,dx=\frac{4}{6}\sin(6x)\)
   • \(-\frac{5}{3}\cos(3x)+\frac{2}{3}\sin(6x)+c\)

3. (c) \(\int 4x(x^2+2)^{6}\,dx\)

   • \(u=x^2+2,\ du=2x\,dx\), so \(4x\,dx=2\,du\)
   • \(\int 2u^{6}\,du=\frac{2}{7}u^{7}\)
   • \(\frac{2}{7}(x^2+2)^{7}+c\)

Find each integral.

1. (a) \(\int x^2\sqrt{3x^3+7}\,dx\)

   • \(u=3x^3+7,\ du=9x^2\,dx\), so \(x^2\,dx=\frac{1}{9}du\)
   • \(\int\frac{1}{9}u^{1/2}\,du=\frac{1}{9}\cdot\frac{2}{3}u^{3/2}\)
   • \(=\frac{2}{27}(3x^3+7)^{3/2}+c\)

2. (b) \(\int x\sqrt[3]{5-3x^2}\,dx\)

   • \(u=5-3x^2,\ du=-6x\,dx\), so \(x\,dx=-\frac{1}{6}du\)
   • \(\int -\frac{1}{6}u^{1/3}\,du=-\frac{1}{6}\cdot\frac{3}{4}u^{4/3}\)
   • \(=-\frac{1}{8}(5-3x^2)^{4/3}+c\)

Find each integral.

1. (a) \(\int 14xe^{4x^2}\,dx\)

   • \(u=4x^2,\ du=8x\,dx\)
   • \(\int 14xe^{u}\,dx=\frac{7}{4}\int e^u\,du\) (since \(14x\,dx=\frac{7}{4}\,du\))
   • \(=\frac{7}{4}e^{4x^2}+c\)

2. (b) \(\int 9x^4e^{-3x^5}\,dx\)

   • \(u=-3x^5,\ du=-15x^4\,dx\)
   • \(\int 9x^4e^{u}\,dx=-\frac{3}{5}\int e^u\,du\) (since \(9x^4\,dx=-\frac{3}{5}\,du\))
   • \(=-\frac{3}{5}e^{-3x^5}+c\)

Find each integral.

1. (a) \(\int\frac{x^4}{3x^5+2}\,dx\)

   • \(u=3x^5+2,\ du=15x^4\,dx\)
   • \(\int\frac{1}{u}\cdot\frac{1}{15}du=\frac{1}{15}\ln|u|+c\)
   • \(\frac{1}{15}\ln|3x^5+2|+c\)

2. (b) \(\int\frac{x}{5-2x^2}\,dx\)

   • \(u=5-2x^2,\ du=-4x\,dx\)
   • \(\int\frac{1}{u}\cdot\left(-\frac{1}{4}\right)du=-\frac{1}{4}\ln|u|+c\)
   • \(-\frac{1}{4}\ln|5-2x^2|+c\)

Use a substitution to find the following.

1. (a) \(\frac{dy}{dx}=x\cos(4x^2)\). Find y in terms of x.

   • \(u=4x^2,\ du=8x\,dx\), so \(x\,dx=\frac{1}{8}du\)
   • \(y=\frac{1}{8}\int \cos u\,du=\frac{1}{8}(\sin u)+c\)
   • \(y=\frac{1}{8}\sin(4x^2)+c\)

2. (b) \(\int\frac{6x}{\sqrt{3x^2+1}}\,dx\)

   • \(u=3x^2+1,\ du=6x\,dx\), so \(6x\,dx=du\)
   • \(\int u^{-1/2}\,du=2u^{1/2}\)
   • \(2\sqrt{3x^2+1}+c\)

Air is pumped into a balloon. After t seconds the volume of air is V cm³, and \(\frac{dV}{dt}=0.9t^2+3\) and V = 20 when t = 0.

1. (a) Find an expression for \(V\) in terms of t.

   • \(V=\int(0.9t^2+3)\,dt=0.3t^3+3t+c\)
   • \(t=0: 20=c\)
   • \(V=0.3t^3+3t+20\)

2. (b) Find the volume when t = 4.

   • \(V(4)=0.3(4)^3+3(4)+20\)
   • = 51.2 cm³

The depth of a submarine below the surface is x metres at time t minutes, and \(\frac{dx}{dt}=15-4e^t\) and x = 30 when t = 0.

1. (a) Find an expression for \(x\) in terms of t.

   • \(x=\int(15-4e^t)\,dt=15t-4e^t+c\)
   • \(t=0: 30=-4+c\)
   • \(c=34\)
   • \(x=15t-4e^t+34\)

2. (b) Find the depth when t = 2.

   • \(x(2)=15(2)-4e^{2}+34\)
   • = 34.4 metres

Find f(x) in each case.

1. (a) \(f'(x)=\frac{4}{(3x-2)^2}\), and the graph of y = f(x) passes through the point (2, 1). Find f(x).

   • \(f(x)=\int 4(3x-2)^{-2}\,dx=\frac{4}{3}\cdot\frac{(3x-2)^{-1}}{-1}+c\) (reverse chain rule: divide by the 'inside' derivative 3)
   • \(f(x)=-\frac{4}{3(3x-2)}+c\)
   • \(f(2)=-\frac{4}{3\times 4}+c=1\), so \(c=\frac{4}{3}\)
   • \(f(x)=-\frac{4}{3(3x-2)}+\frac{4}{3}\)

2. (b) \(f'(x)=\frac{3\ln x}{4x}\) for x > 0, and the curve y = f(x) passes through the point \((e^2, 6)\). Find f(x).

   • \(u=\ln x,\ du=\frac{1}{x}dx\), so \(f'(x)\,dx=\frac{3}{4}u\,du\)
   • \(f(x)=\int \frac{3}{4}u\,du=\frac{3}{8}u^2+c=\frac{3}{8}(\ln x)^2+c\)
   • \(f(e^2)=\frac{3}{8}\times 2^2+c=6\), so \(c=\frac{9}{2}\)
   • \(f(x)=\frac{3}{8}(\ln x)^2+\frac{9}{2}\)

\(f'(x)=\frac{5}{x}\) and f(-3) = 2. Find the exact value of \(f(-7)\).

• \(f(x)=5\ln|x|+c\) (x is negative, so use the absolute value)
• \(f(-3)=5\ln 3+c=2\), so \(c=2-5\ln 3\)
• \(f(-7)=5\ln 7+2-5\ln 3\)
• \(=2+5\ln\frac{7}{3}\), which is 6.24 (3 s.f.)

Find each integral.

1. (a) \(\int\left(10\cos^4 x\sin x+kx^{2}+3\right)dx\), where k is a constant.

   • \(u=\cos x,\ du=-\sin x\,dx\): \(\int 10u^4(-du)=-2u^5\)
   • \(\int kx^{2}\,dx=\frac{kx^{3}}{3}\) (k is a constant)
   • \(\int 3\,dx=3x\)
   • \(-2\cos^5 x+\frac{k x^{3}}{3}+3x+c\)

2. (b) \(\int\frac{e^{3x}+e^{-x}}{e^{x}}\,dx\)

   • \(\frac{e^{3x}+e^{-x}}{e^{x}}=e^{2x}+e^{-2x}\) (subtract the powers)
   • \(\int\left(e^{2x}+e^{-2x}\right)dx\) , divide each by the coefficient of x
   • \(\frac{1}{2}e^{2x}-\frac{1}{2}e^{-2x}+c\)

Find each integral.

1. (a) \(\int\sqrt{x^4+8x^2}\,dx\) for x > 0.

   • \(\sqrt{x^4+8x^2}=\sqrt{x^2(x^2+8)}=x\sqrt{x^2+8}\) (x > 0, so \(\sqrt{x^2}=x\))
   • \(u=x^2+8,\ du=2x\,dx\): \(\int\frac{1}{2}u^{1/2}du=\frac{1}{3}u^{3/2}\)
   • \(\frac{1}{3}(x^2+8)^{3/2}+c\)

2. (b) \(\int\frac{4}{x\ln x}\,dx\) for x > 1.

   • \(u=\ln x,\ du=\frac{1}{x}dx\): the integral is \(\int\frac{4}{u}\,du\)
   • \(=4\ln|u|+c=4\ln|\ln x|+c\)

Find each integral.

1. (a) Given that \(\cos 2x=1-2\sin^2x\), find \(\int 8\sin^2x\,dx\).

   • \(\sin^2x=\frac{1-\cos 2x}{2}\) (rearrange the identity)
   • \(\int 8\sin^2x\,dx=4\int(1-\cos 2x)\,dx\)
   • \(4x-2\sin 2x+c\)

2. (b) \(\int\tan 4x\,dx\), showing your working.

   • \(\tan 4x=\frac{\sin 4x}{\cos 4x}\)
   • \(u=\cos 4x,\ du=-4\sin 4x\,dx\), so \(\sin 4x\,dx=-\frac{1}{4}du\)
   • \(\int\frac{1}{u}\left(-\frac{1}{4}\right)du=-\frac{1}{4}\ln|u|+c\)
   • \(-\frac{1}{4}\ln|\cos 4x|+c\)

3. (c) \(\int 3\cos^3x\,dx\)

   • \(3\cos^3x=3\cos x(1-\sin^2x)\) (use \(\cos^2x=1-\sin^2x\))
   • \(u=\sin x,\ du=\cos x\,dx\): \(\int 3(1-u^2)\,du\)
   • \(=3u-u^3+c\)
   • \(3\sin x-\sin^3x+c\)