Integration techniques and differential equations – Year 13 A level maths revision notes to download as PDF

Worked example 1

Evaluate \(\int_0^1e^{2x}dx\) and find \(\int\dfrac{1}{2x+1}dx\), \(\int\tan x\,dx\) and \(\int\sin^2x\,dx\).

\(\left[\dfrac12e^{2x}\right]_0^1=\dfrac{e^2-1}{2}\approx3.19\). Then \(\tfrac12\ln|2x+1|+c\) and \(-\ln|\cos x|+c\). For \(\sin^2x=\dfrac{1-\cos2x}2\): \(\dfrac x2-\dfrac{\sin2x}4+c\).

Worked example 2

Evaluate \(\lim_{\delta x\to0}\sum_{x=1}^{4}x^2\delta x\).

It equals \(\int_1^4x^2dx=\left[\dfrac{x^3}3\right]_1^4=21\).

Worked example 3

Find \(\int2x(x^2+1)^3dx\) and \(\int_0^1x\sqrt{1+x^2}\,dx\).

\(u=x^2+1\), \(du=2x\,dx\): \(\int u^3du=\dfrac{(x^2+1)^4}{4}+c\). For the second \(u=1+x^2\) gives \(\tfrac12\int_1^2u^{1/2}du=\dfrac{2\sqrt2-1}3\approx0.609\).

Worked example 4

Evaluate \(\int_0^3\dfrac{x}{\sqrt{x+1}}dx\).

Let \(u=x+1\), \(x=u-1\), \(dx=du\). Limits: \(x=0\Rightarrow u=1\); \(x=3\Rightarrow u=4\). The integral is \(\int_1^4\left(u^{1/2}-u^{-1/2}\right)du=\left[\dfrac23u^{3/2}-2u^{1/2}\right]_1^4=\left(\dfrac{16}3-4\right)-\left(\dfrac23-2\right)=\dfrac83\).

Worked example 5

Find \(\int xe^xdx\) and \(\int_0^\pi x\sin x\,dx\).

With \(u=x\), \(\dfrac{dv}{dx}=e^x\): \(xe^x-\int e^xdx=(x-1)e^x+c\). For the second, \(u=x\), \(v=-\cos x\): \([-x\cos x]_0^\pi+\int_0^\pi\cos x\,dx=\pi+0=\pi\).

Worked example 6

Find \(\int\ln x\,dx\) and \(\int x^2e^xdx\).

Take \(u=\ln x\), \(\dfrac{dv}{dx}=1\): \(x\ln x-\int1\,dx=x\ln x-x+c\), so \(\int_1^e\ln x\,dx=1\). Repeating the method gives \(\int x^2e^xdx=e^x(x^2-2x+2)+c\).

Worked example 7

Evaluate \(\int_2^3\dfrac{5x+1}{(x-1)(x+2)}dx\).

\(\dfrac{5x+1}{(x-1)(x+2)}=\dfrac2{x-1}+\dfrac3{x+2}\), so the integral is \(\left[2\ln|x-1|+3\ln|x+2|\right]_2^3=2\ln2+3\ln5-3\ln4=3\ln5-4\ln2\approx2.056\). Use the laws of logarithms for a simple exact answer.

Worked example 8

Solve \(\dfrac{dy}{dx}=2xy\) with \(y=3\) when \(x=0\).

\(\int\dfrac1ydy=\int2x\,dx\), so \(\ln y=x^2+c\), \(y=Ae^{x^2}\), and \(A=3\): \(y=3e^{x^2}\).

Worked example 9

A population satisfies \(\dfrac{dP}{dt}=0.05P\). When does it double?

\(P=P_0e^{0.05t}\), and \(e^{0.05t}=2\) gives \(t=20\ln2\approx13.9\). (Newton's law of cooling \(\dfrac{d\theta}{dt}=-k(\theta-20)\) gives \(\theta-20=Ae^{-kt}\).)

Worked example 10

A tank empties with \(\dfrac{dh}{dt}=-0.2\sqrt h\), depth \(h\) m, \(t\) in minutes, \(h=4\) at \(t=0\). Find \(h\) and the time to empty.

\(\int h^{-1/2}dh=\int-0.2\,dt\), so \(2\sqrt h=-0.2t+c\). With \(h=4\) at \(t=0\), \(c=4\). Then \(\sqrt h=2-0.1t\) and \(h=(2-0.1t)^2\). The tank is empty after 20 minutes. The formula is only valid for \(0\le t\le20\): afterwards it would give the depth rising again, which is impossible.

Worked example 11: A sum that becomes an integral

Write \(\lim_{\delta x\to0}\sum_{x=0}^{2}\left(3x^2+1\right)\delta x\) as an integral and evaluate it.

The sum of thin rectangles of height \(3x^2+1\) and width \(\delta x\) becomes \(\int_0^2(3x^2+1)dx=\left[x^3+x\right]_0^2=8+2=10\). The sigma notation is just a way of saying that we add up the areas of the strips.

Worked example 12: A trigonometric integral using an identity

Find \(\int\cos^2x\,dx\) and evaluate \(\int_0^{\pi/2}\cos^2x\,dx\).

Use \(\cos^2x=\dfrac{1+\cos2x}{2}\). Then \(\int\cos^2x\,dx=\dfrac x2+\dfrac{\sin2x}4+c\). Between \(0\) and \(\tfrac\pi2\): \(\dfrac\pi4+0-0=\dfrac\pi4\). Notice that you cannot integrate \(\cos^2x\) directly; rewrite it first.

Worked example 13: Spotting a reverse chain rule

Evaluate \(\int_0^{\pi/4}\tan x\sec^2x\,dx\).

Let \(u=\tan x\), so \(du=\sec^2x\,dx\). Limits: \(x=0\Rightarrow u=0\); \(x=\tfrac\pi4\Rightarrow u=1\). \(\int_0^1u\,du=\tfrac12\). The integrand is a function (\(\tan x\)) multiplied by its own derivative (\(\sec^2x\)), which is the signal for substitution.

Worked example 14: Parts applied to a trigonometric product

Find \(\int x\cos2x\,dx\).

\(u=x\), \(\dfrac{dv}{dx}=\cos2x\), \(v=\tfrac12\sin2x\). \(\int x\cos2x\,dx=\tfrac12x\sin2x-\tfrac12\int\sin2x\,dx=\tfrac12x\sin2x+\tfrac14\cos2x+c\). Check by differentiating: \(\tfrac12\sin2x+x\cos2x-\tfrac12\sin2x=x\cos2x\).

Worked example 15: An exact answer using partial fractions and logarithms

Show that \(\int_1^2\dfrac{4}{(x+1)(x+3)}dx=2\ln\dfrac65\).

\(\dfrac4{(x+1)(x+3)}=\dfrac2{x+1}-\dfrac2{x+3}\). Integrating: \(\left[2\ln\dfrac{x+1}{x+3}\right]_1^2=2\ln\dfrac35-2\ln\dfrac24=2\ln\dfrac{3/5}{1/2}=2\ln\dfrac65=0.365\). Combining logarithms before substituting the limits makes the algebra shorter.

Worked example 16: Newton law of cooling with a negative temperature drop

A metal bar at \(200^\circ\)C is placed in a room at \(20^\circ\)C. Given that \(\dfrac{d\theta}{dt}=-0.05(\theta-20)\), find the time for the bar to cool to \(100^\circ\)C.

The solution is \(\theta=20+180e^{-0.05t}\). For \(\theta=100\): \(e^{-0.05t}=\dfrac{80}{180}=\dfrac49\), so \(t=\dfrac{\ln(9/4)}{0.05}=16.2\) minutes.

Worked example 17: Parts with a logarithm and a definite integral

Evaluate \(\int_1^4\dfrac{\ln x}{\sqrt x}dx\).

Take \(u=\ln x\), \(\dfrac{dv}{dx}=x^{-1/2}\), so \(v=2\sqrt x\). Then \(\int\dfrac{\ln x}{\sqrt x}dx=2\sqrt x\ln x-\int2\sqrt x\cdot\dfrac1xdx=2\sqrt x\ln x-\int2x^{-1/2}dx=2\sqrt x\ln x-4\sqrt x\). Between 1 and 4: \((4\ln4-8)-(0-4)=4\ln4-4=8\ln2-4=1.55\). Checking by differentiating the antiderivative confirms the result.

Worked example 18: A definite integral with substitution and exponential

Use \(u=e^x\) to evaluate \(\int_0^{\ln3}\dfrac{e^x}{1+e^{2x}}dx\).

With \(u=e^x\), \(du=e^xdx\). Limits: \(x=0\Rightarrow u=1\); \(x=\ln3\Rightarrow u=3\). The integral becomes \(\int_1^3\dfrac{1}{1+u^2}du=\left[\tan^{-1}u\right]_1^3=\tan^{-1}3-\dfrac\pi4=0.4636\). Recognising \(\dfrac{1}{1+u^2}\) as the derivative of \(\tan^{-1}u\) is the key step.

Worked example 19: Mean value of a function

The mean value of \(f(x)\) over \(a\le x\le b\) is \(\dfrac{1}{b-a}\int_a^bf(x)dx\). Find the mean value of \(f(x)=e^{-x}\) over \(0\le x\le2\).

\(\int_0^2e^{-x}dx=\left[-e^{-x}\right]_0^2=1-e^{-2}\). The width is 2, so the mean value is \(\dfrac{1-e^{-2}}{2}=0.432\). This is the height of a rectangle on the same base with the same area as the region under the curve.

Worked example 20: Two stages: separating the variables with a trigonometric function

Solve \(\dfrac{dy}{dx}=y\cos x\) with \(y=2\) when \(x=0\).

\(\int\dfrac1ydy=\int\cos x\,dx\), so \(\ln y=\sin x+c\). With \(y=2\) at \(x=0\), \(c=\ln2\), so \(\ln y-\ln2=\sin x\) and \(y=2e^{\sin x}\). The solution oscillates between \(2e^{-1}=0.74\) and \(2e=5.44\).

Worked example 21: a leaking water tank

Water leaks from a tank so that \(\dfrac{dh}{dt}=-k\sqrt h\), where \(h\) m is the depth after \(t\) minutes. The depth is 4 m at the start and 1 m after 10 minutes. Find \(h\) in terms of \(t\) and the time when the tank is empty.

Separate the variables: \(\int h^{-1/2}dh=\int-k\,dt\), so \(2\sqrt h=-kt+c\). At \(t=0\), \(h=4\), so \(c=2\times2=4\). At \(t=10\), \(h=1\): \(2=-10k+4\), so \(k=0.2\). Hence \(2\sqrt h=4-0.2t\), which gives \(h=\left(2-0.1t\right)^2\). The tank is empty when \(2-0.1t=0\), that is after \(t=20\) minutes. The model is only valid for \(0\le t\le20\); after that the formula would wrongly predict the depth rising again.