Unit 27 Review Exercises

Integration Techniques

These questions mix the ideas from the whole unit. Paper 1 means no calculator; Paper 2 means a GDC is allowed. AHL marks questions for HL students only.

1 Paper 1 A curve has gradient \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6x - \dfrac{2}{x^2}\) and passes through the point \((1, 7)\). Find the equation of the curve.

Solution

\(y = \displaystyle\int \left(6x - 2x^{-2}\right)\mathrm{d}x = 3x^2 + \frac{2}{x} + C\).

At \((1, 7)\): \(7 = 3 + 2 + C\), so \(C = 2\). \(\;y = 3x^2 + \dfrac{2}{x} + 2\). See 27.1.

2 Paper 1 Find \(\displaystyle\int \left(\mathrm{e}^{2x} + \frac{1}{3x + 1}\right)\mathrm{d}x\).

Solution

\(\dfrac{1}{2}\mathrm{e}^{2x} + \dfrac{1}{3}\ln\lvert 3x + 1\rvert + C\). See 27.2.

3 Paper 1 Find \(\displaystyle\int \sin^2 4x\,\mathrm{d}x\).

Solution

\(\sin^2 4x = \dfrac{1}{2}(1 - \cos 8x)\), so the integral is \(\dfrac{x}{2} - \dfrac{\sin 8x}{16} + C\). See 27.3.

4 Paper 1 Find \(\displaystyle\int 5x^2\left(2x^3 - 1\right)^4\mathrm{d}x\).

Solution

\(g(x) = 2x^3 - 1\), \(g'(x) = 6x^2\), so \(5x^2 = \dfrac{5}{6}g'(x)\).

\(\dfrac{5}{6} \cdot \dfrac{\left(2x^3 - 1\right)^5}{5} + C = \dfrac{\left(2x^3 - 1\right)^5}{6} + C\). See 27.2 and 27.5.

5 Paper 1 AHL

(a) Express \(\dfrac{x + 7}{x^2 - x - 6}\) in partial fractions.

(b) Hence find \(\displaystyle\int \frac{x + 7}{x^2 - x - 6}\,\mathrm{d}x\).

Solution

(a) \(x^2 - x - 6 = (x - 3)(x + 2)\). \(\;x + 7 = A(x + 2) + B(x - 3)\). \(x = 3\): \(10 = 5A\), \(A = 2\). \(\;x = -2\): \(5 = -5B\), \(B = -1\). \(\;\dfrac{2}{x - 3} - \dfrac{1}{x + 2}\).

(b) \(2\ln\lvert x - 3\rvert - \ln\lvert x + 2\rvert + C\). See 27.4.

6 Paper 1 AHL Find \(\displaystyle\int \frac{1}{x^2 - 6x + 13}\,\mathrm{d}x\).

Solution

\(x^2 - 6x + 13 = (x - 3)^2 + 4\): \(\;\dfrac{1}{2}\arctan\left(\dfrac{x - 3}{2}\right) + C\). See 27.4.

7 Paper 1 AHL Find \(\displaystyle\int x\cos 3x\,\mathrm{d}x\).

Solution

Parts, \(u = x\), \(v = \dfrac{1}{3}\sin 3x\): \(\;\dfrac{x}{3}\sin 3x - \displaystyle\int \frac{1}{3}\sin 3x\,\mathrm{d}x = \frac{x}{3}\sin 3x + \frac{1}{9}\cos 3x + C\). See 27.6.

8 Paper 1 AHL By using the substitution \(u = x - 1\), find \(\displaystyle\int x\sqrt{x - 1}\,\mathrm{d}x\).

Solution

\(x = u + 1\), \(\mathrm{d}x = \mathrm{d}u\): \(\;\displaystyle\int (u + 1)u^{\frac{1}{2}}\,\mathrm{d}u = \int \left(u^{\frac{3}{2}} + u^{\frac{1}{2}}\right)\mathrm{d}u = \frac{2}{5}u^{\frac{5}{2}} + \frac{2}{3}u^{\frac{3}{2}} + C\).

\(= \dfrac{2}{5}(x - 1)^{\frac{5}{2}} + \dfrac{2}{3}(x - 1)^{\frac{3}{2}} + C\). See 27.5.

9 Paper 1 AHL Find \(\displaystyle\int \mathrm{e}^x\cos x\,\mathrm{d}x\).

Solution

Let \(I\) be the integral. \(u = \cos x\): \(\;I = \mathrm{e}^x\cos x + \displaystyle\int \mathrm{e}^x\sin x\,\mathrm{d}x\). \(u = \sin x\): \(\;\displaystyle\int \mathrm{e}^x\sin x\,\mathrm{d}x = \mathrm{e}^x\sin x - I\).

So \(2I = \mathrm{e}^x(\cos x + \sin x)\) and \(I = \dfrac{1}{2}\mathrm{e}^x(\sin x + \cos x) + C\). See 27.6.

10 Paper 2 A function has \(f'(x) = \mathrm{e}^{-x^2}\) and \(f(0) = 2\).

(a) Explain why \(f(1) = 2 + \displaystyle\int_0^1 \mathrm{e}^{-x^2}\,\mathrm{d}x\).

(b) Find \(f(1)\).

Solution

(a) \(\mathrm{e}^{-x^2}\) has no simple anti-derivative (it is not a standard form, and \(-2x\) is missing for inspection). But \(f\) is an anti-derivative of \(f'\), so \(\displaystyle\int_0^1 f'(x)\,\mathrm{d}x = f(1) - f(0)\), which gives the result.

(b) Calculator → Menu → Calculus → Numerical Integral: \(\displaystyle\int_0^1 \mathrm{e}^{-x^2}\,\mathrm{d}x = 0.746\,82\ldots\)

\(f(1) = 2.746\,82\ldots \approx 2.75\) (3 s.f.). See 27.1 and Unit 28.