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IB Mathematics: Applications and Interpretation HL · AHL 3: Further Integration · AHL 3.3

Further integration: practice Problem set A Worked solutions

Skills

  1. \(\frac{dy}{dx}=15\sqrt{x}\), and y = 75 when x = 4. Find y in terms of x.

    • \(y=\int 15x^{1/2}\,dx=10x^{3/2}+c\)
    • \(x=4, y=75: 75=10(4)^{3/2}+c=80+c\)
    • \(c=-5\)
    • \(y=10x^{3/2}-5\)

  2. Find the required function in each case.

    1. (a) \(f'(x)=6\cos x-5\sin x\) and f(0) = 9. Find f(x).

      • \(f(x)=6\sin x+5\cos x+c\) (the integral of \(-\sin x\) is \(\cos x\))
      • \(f(0)=5+c=9\)
      • \(c=4\)
      • \(f(x)=6\sin x+5\cos x+4\)

    2. (b) \(\frac{dy}{dx}=3e^x+\frac{4}{x}\) for x > 0, and y = 2 when x = 1. Find y in terms of x.

      • \(y=3e^x+4\ln x+c\)
      • \(x=1: 2=3e+4\ln 1+c=3e+c\)
      • \(c=2-3e\)
      • \(y=3e^x+4\ln x+2-3e\), which is \(y=3e^x+4\ln x-6.15\) (3 s.f.)

    3. (c) \(\int\frac{9x^2+4}{3x}\,dx\)

      • \(\frac{9x^2+4}{3x}=3x+\frac{4}{3x}\) (divide each term by the denominator)
      • \(\int\left(3x+\frac{4}{3x}\right)dx=\frac{3}{2}x^2+\frac{4}{3}\ln|x|+c\)

  3. Find each integral.

    1. (a) \(\int(7-3x)^{8}\,dx\)

      • \(u=7-3x,\ \frac{du}{dx}=-3\), so \(dx=-\frac{1}{3}du\)
      • \(\int u^{8}\left(-\frac{1}{3}\right)du=-\frac{1}{3}\cdot\frac{u^{9}}{9}\)
      • \(=-\frac{1}{27}(7-3x)^{9}+c\)

    2. (b) \(\int(3\sin(4x)+8\cos(2x))\,dx\)

      • \(\int 3\sin(4x)\,dx=-\frac{3}{4}\cos(4x)\) (divide by the coefficient of x)
      • \(\int 8\cos(2x)\,dx=\frac{8}{2}\sin(2x)\)
      • \(-\frac{3}{4}\cos(4x)+4\sin(2x)+c\)

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

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

  4. Find each integral.

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

      • \(u=5x^3+2,\ du=15x^2\,dx\), so \(x^2\,dx=\frac{1}{15}du\)
      • \(\int\frac{1}{15}u^{1/2}\,du=\frac{1}{15}\cdot\frac{2}{3}u^{3/2}\)
      • \(=\frac{2}{45}(5x^3+2)^{3/2}+c\)

    2. (b) \(\int x\sqrt[3]{7-2x^2}\,dx\)

      • \(u=7-2x^2,\ du=-4x\,dx\), so \(x\,dx=-\frac{1}{4}du\)
      • \(\int -\frac{1}{4}u^{1/3}\,du=-\frac{1}{4}\cdot\frac{3}{4}u^{4/3}\)
      • \(=-\frac{3}{16}(7-2x^2)^{4/3}+c\)

  5. Find each integral.

    1. (a) \(\int 9xe^{6x^2}\,dx\)

      • \(u=6x^2,\ du=12x\,dx\)
      • \(\int 9xe^{u}\,dx=\frac{3}{4}\int e^u\,du\) (since \(9x\,dx=\frac{3}{4}\,du\))
      • \(=\frac{3}{4}e^{6x^2}+c\)

    2. (b) \(\int 8x^2e^{-2x^3}\,dx\)

      • \(u=-2x^3,\ du=-6x^2\,dx\)
      • \(\int 8x^2e^{u}\,dx=-\frac{4}{3}\int e^u\,du\) (since \(8x^2\,dx=-\frac{4}{3}\,du\))
      • \(=-\frac{4}{3}e^{-2x^3}+c\)

  6. Find each integral.

    1. (a) \(\int\frac{x^2}{2x^3+9}\,dx\)

      • \(u=2x^3+9,\ du=6x^2\,dx\)
      • \(\int\frac{1}{u}\cdot\frac{1}{6}du=\frac{1}{6}\ln|u|+c\)
      • \(\frac{1}{6}\ln|2x^3+9|+c\)

    2. (b) \(\int\frac{x^3}{1-x^4}\,dx\)

      • \(u=1-x^4,\ du=-4x^3\,dx\)
      • \(\int\frac{1}{u}\cdot\left(-\frac{1}{4}\right)du=-\frac{1}{4}\ln|u|+c\)
      • \(-\frac{1}{4}\ln|1-x^4|+c\)

  7. Use a substitution to find the following.

    1. (a) \(\frac{dy}{dx}=x\sin(5x^2)\). Find y in terms of x.

      • \(u=5x^2,\ du=10x\,dx\), so \(x\,dx=\frac{1}{10}du\)
      • \(y=\frac{1}{10}\int \sin u\,du=\frac{1}{10}(-\cos u)+c\)
      • \(y=-\frac{1}{10}\cos(5x^2)+c\)

    2. (b) \(\int\frac{4x}{\sqrt{2x^2+9}}\,dx\)

      • \(u=2x^2+9,\ du=4x\,dx\), so \(4x\,dx=du\)
      • \(\int u^{-1/2}\,du=2u^{1/2}\)
      • \(2\sqrt{2x^2+9}+c\)

In context

  1. Water flows into a tank. After t minutes the volume of water is V litres, and \(\frac{dV}{dt}=0.6t^2+2\) and V = 15 when t = 0.

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

      • \(V=\int(0.6t^2+2)\,dt=0.2t^3+2t+c\)
      • \(t=0: 15=c\)
      • \(V=0.2t^3+2t+15\)

    2. (b) Find the volume when t = 5.

      • \(V(5)=0.2(5)^3+2(5)+15\)
      • = 50 litres

  2. The height of a drone above the ground is x metres at time t seconds, and \(\frac{dx}{dt}=12-5e^t\) and x = 8 when t = 0.

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

      • \(x=\int(12-5e^t)\,dt=12t-5e^t+c\)
      • \(t=0: 8=-5+c\)
      • \(c=13\)
      • \(x=12t-5e^t+13\)

    2. (b) Find the height when t = 1.5.

      • \(x(1.5)=12(1.5)-5e^{1.5}+13\)
      • = 8.59 metres

Stretch

  1. Find f(x) in each case.

    1. (a) \(f'(x)=\frac{3}{(2x+1)^2}\), and the graph of y = f(x) passes through the point (1, 0). Find f(x).

      • \(f(x)=\int 3(2x+1)^{-2}\,dx=\frac{3}{2}\cdot\frac{(2x+1)^{-1}}{-1}+c\) (reverse chain rule: divide by the 'inside' derivative 2)
      • \(f(x)=-\frac{3}{2(2x+1)}+c\)
      • \(f(1)=-\frac{3}{2\times 3}+c=0\), so \(c=\frac{1}{2}\)
      • \(f(x)=-\frac{3}{2(2x+1)}+\frac{1}{2}\)

    2. (b) \(f'(x)=\frac{5\ln x}{2x}\) for x > 0, and the curve y = f(x) passes through the point \((e, 3)\). Find f(x).

      • \(u=\ln x,\ du=\frac{1}{x}dx\), so \(f'(x)\,dx=\frac{5}{2}u\,du\)
      • \(f(x)=\int \frac{5}{2}u\,du=\frac{5}{4}u^2+c=\frac{5}{4}(\ln x)^2+c\)
      • \(f(e)=\frac{5}{4}\times 1^2+c=3\), so \(c=\frac{7}{4}\)
      • \(f(x)=\frac{5}{4}(\ln x)^2+\frac{7}{4}\)

  2. \(f'(x)=\frac{3}{x}\) and f(-2) = 4. Find the exact value of \(f(-5)\).

    • \(f(x)=3\ln|x|+c\) (x is negative, so use the absolute value)
    • \(f(-2)=3\ln 2+c=4\), so \(c=4-3\ln 2\)
    • \(f(-5)=3\ln 5+4-3\ln 2\)
    • \(=4+3\ln\frac{5}{2}\), which is 6.75 (3 s.f.)

  3. Find each integral.

    1. (a) \(\int\left(8\sin^3 x\cos x+kx^{3}+2\right)dx\), where k is a constant.

      • \(u=\sin x,\ du=\cos x\,dx\): \(\int 8u^3\,du=2u^4\)
      • \(\int kx^{3}\,dx=\frac{kx^{4}}{4}\) (k is a constant)
      • \(\int 2\,dx=2x\)
      • \(2\sin^4 x+\frac{k x^{4}}{4}+2x+c\)

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

      • \(\frac{e^{5x}-e^{-x}}{e^{2x}}=e^{3x}-e^{-3x}\) (subtract the powers)
      • \(\int\left(e^{3x}-e^{-3x}\right)dx\) , divide each by the coefficient of x
      • \(\frac{1}{3}e^{3x}+\frac{1}{3}e^{-3x}+c\)

  4. Find each integral.

    1. (a) \(\int\sqrt{x^4+5x^2}\,dx\) for x > 0.

      • \(\sqrt{x^4+5x^2}=\sqrt{x^2(x^2+5)}=x\sqrt{x^2+5}\) (x > 0, so \(\sqrt{x^2}=x\))
      • \(u=x^2+5,\ du=2x\,dx\): \(\int\frac{1}{2}u^{1/2}du=\frac{1}{3}u^{3/2}\)
      • \(\frac{1}{3}(x^2+5)^{3/2}+c\)

    2. (b) \(\int\frac{6}{x\ln x}\,dx\) for x > 1.

      • \(u=\ln x,\ du=\frac{1}{x}dx\): the integral is \(\int\frac{6}{u}\,du\)
      • \(=6\ln|u|+c=6\ln|\ln x|+c\)

  5. Find each integral.

    1. (a) Given that \(\cos 2x=2\cos^2x-1\), find \(\int 6\cos^2x\,dx\).

      • \(\cos^2x=\frac{1+\cos 2x}{2}\) (rearrange the identity)
      • \(\int 6\cos^2x\,dx=3\int(1+\cos 2x)\,dx\)
      • \(3x+\frac{3}{2}\sin 2x+c\)

    2. (b) \(\int\tan 3x\,dx\), showing your working.

      • \(\tan 3x=\frac{\sin 3x}{\cos 3x}\)
      • \(u=\cos 3x,\ du=-3\sin 3x\,dx\), so \(\sin 3x\,dx=-\frac{1}{3}du\)
      • \(\int\frac{1}{u}\left(-\frac{1}{3}\right)du=-\frac{1}{3}\ln|u|+c\)
      • \(-\frac{1}{3}\ln|\cos 3x|+c\)

    3. (c) \(\int 6\sin^3x\,dx\)

      • \(6\sin^3x=6\sin x(1-\cos^2x)\) (use \(\sin^2x=1-\cos^2x\))
      • \(u=\cos x,\ du=-\sin x\,dx\): \(\int 6(1-u^2)(-du)=6\int(u^2-1)\,du\)
      • \(=2u^3-6u+c\)
      • \(-6\cos x+2\cos^3x+c\)

jmaths.xyz · AI HL · AHL 3.3 Further integration: substitution and boundary conditions