Integration – Year 12 A level maths revision notes to download as PDF

Worked example 1

Find \(\displaystyle\int(6x^2-4x+3)\,dx\) and \(\displaystyle\int\left(\sqrt x+\frac3{x^2}\right)dx\).

The first is \(2x^3-2x^2+3x+c\). For the second, \(\displaystyle\int\left(x^{1/2}+3x^{-2}\right)dx=\frac23x^{3/2}-3x^{-1}+c\).

Worked example 2

Find \(\displaystyle\int\dfrac{x^2+5}{x^2}\,dx\).

Divide first: \(\displaystyle\int(1+5x^{-2})\,dx=x-\dfrac5x+c\).

Worked example 3

Given \(\dfrac{dy}{dx}=3x^2-4\) and that the curve passes through \((2,5)\), find \(y\).

\(y=x^3-4x+c\), so \(5=8-8+c\) and \(c=5\). The curve is \(y=x^3-4x+5\).

Worked example 4

Given \(f'(x)=3x^2-6x\) and \(f(1)=0\), find \(f(2)\).

\(f(x)=x^3-3x^2+c\). Then \(f(1)=1-3+c=0\), so \(c=2\) and \(f(x)=x^3-3x^2+2\). So \(f(2)=8-12+2=-2\).

Worked example 5

Evaluate \(\displaystyle\int_1^3(2x+1)\,dx\).

\(\Big[x^2+x\Big]_1^3=(9+3)-(1+1)=10\).

Worked example 6

Find \(k>0\) such that \(\displaystyle\int_1^k2x\,dx=8\).

\(\big[x^2\big]_1^k=k^2-1=8\), so \(k^2=9\) and \(k=3\) (rejecting \(k=-3\) because \(k>0\)).

Worked example 7

Find the area under \(y=x^2+1\) from \(x=0\) to \(x=3\).

\(\displaystyle\int_0^3(x^2+1)\,dx=\Big[\tfrac{x^3}3+x\Big]_0^3=9+3=12\).

Worked example 8

Find the area between \(y=x^2-3x\) and the \(x\)-axis from \(x=0\) to \(x=3\).

\(\displaystyle\int_0^3(x^2-3x)\,dx=\Big[\tfrac{x^3}3-\tfrac{3x^2}2\Big]_0^3=9-13.5=-4.5\), so the area is \(4.5\).

Worked example 9

Find the area between \(y=x^2\) and \(y=2x\).

They meet where \(x^2=2x\), so \(x=0,2\). On \([0,2]\), \(2x\ge x^2\), so area \(=\displaystyle\int_0^2(2x-x^2)\,dx=\Big[x^2-\tfrac{x^3}3\Big]_0^2=4-\tfrac83=\dfrac43\).

Worked example 10

Find the area between \(y=x^2-2x+3\) and \(y=x+3\).

\(x^2-2x+3=x+3\Rightarrow x^2-3x=0\), so \(x=0\) or \(3\). The line is above the curve, so area \(=\displaystyle\int_0^3(3x-x^2)\,dx=\Big[\tfrac{3x^2}2-\tfrac{x^3}3\Big]_0^3=13.5-9=4.5\).

Worked example 11: Integrating a quotient and a negative power

Find \(\displaystyle\int\dfrac{x^3-2x^2+5}{x^2}\,dx\) and evaluate it between 1 and 2.

Divide each term by \(x^2\): \(x-2+5x^{-2}\). Integrate: \(\dfrac{x^2}2-2x-5x^{-1}+c\). Between 1 and 2: at \(x=2\): \(2-4-2.5=-4.5\); at \(x=1\): \(0.5-2-5=-6.5\). The definite integral is \(-4.5-(-6.5)=2\). Check by estimating: the integrand at \(x=1.5\) is \(1.5-2+2.22=1.72\), so over a width of 1 the area is roughly 1.7 to 2, in line with the answer of 2.

Worked example 12: Area between a curve and the x-axis with a sign change

Find the total area between \(y=x^3-4x\) and the \(x\)-axis for \(-2\le x\le2\).

The curve is \(x(x-2)(x+2)\), crossing at \(-2\), \(0\) and \(2\). For \(-2\lt x<0\) it is above the axis; for \(0\lt x<2\) below. \(\int_{-2}^0(x^3-4x)dx=\left[\dfrac{x^4}4-2x^2\right]_{-2}^0=0-(4-8)=4\). \(\int_0^2=\left[\dfrac{x^4}4-2x^2\right]_0^2=4-8=-4\). The full integral is 0, but the actual area is \(4+4=8\). This is why you must split the integral where the curve crosses the axis.

Worked example 13: Using a boundary condition twice

\(f''(x)=6x-4\), \(f'(1)=2\) and \(f(1)=3\). Find \(f(x)\).

Integrate once: \(f'(x)=3x^2-4x+c\). With \(f'(1)=2\): \(3-4+c=2\), so \(c=3\) and \(f'(x)=3x^2-4x+3\). Integrate again: \(f(x)=x^3-2x^2+3x+d\). With \(f(1)=3\): \(1-2+3+d=3\), so \(d=1\). Therefore \(f(x)=x^3-2x^2+3x+1\). Check: \(f''=6x-4\) ✓, \(f'(1)=3-4+3=2\) ✓, \(f(1)=1-2+3+1=3\) ✓.

Worked example 14: Area enclosed by a curve and a chord

Find the area enclosed between \(y=x^2-2x+3\) and the line \(y=x+3\).

Equate: \(x^2-2x+3=x+3\Rightarrow x^2-3x=0\), so \(x=0\) and \(x=3\). The line is above the curve between them (test \(x=1\): line 4, curve 2). Area \(=\int_0^3\left[(x+3)-(x^2-2x+3)\right]dx=\int_0^3(3x-x^2)dx=\left[\tfrac32x^2-\tfrac{x^3}3\right]_0^3=13.5-9=4.5\). The same answer as the quadratic-and-chord rule \(\dfrac{|a|(\beta-\alpha)^3}6=\dfrac{27}6=4.5\) ✓.

Worked example 15: Linking integration with a rate of change

Oil leaks from a tank at a rate of \(r=0.4\sqrt t\) litres per minute. Find the total leaked in the first 9 minutes.

Total \(=\int_0^9 0.4t^{1/2}dt=0.4\times\dfrac23\left[t^{3/2}\right]_0^9=\dfrac{0.8}3\times27=7.2\) litres. The integral of a rate over a time interval always gives the total change in the quantity. The same principle gives distance from speed, volume from flow rate and cost from marginal cost.

Worked example 16: Integrating with a context and a boundary condition

The gradient of a hillside profile is \(\dfrac{dy}{dx}=0.3x-0.02x^2\) (heights in tens of metres, distance in hundreds of metres). The path starts at the origin. Find the height profile, the height at \(x=10\) and the point where the path is highest.

Integrate: \(y=0.15x^2-\dfrac{0.02}3x^3+c\). The path starts at the origin so \(c=0\). At \(x=10\): \(y=15-6.67=8.33\), so the height is about 83 metres. The highest point occurs where the gradient is zero: \(0.3x-0.02x^2=x(0.3-0.02x)=0\), so \(x=15\), and the height there is \(0.15\times225-\dfrac{0.02}3\times3375=33.75-22.5=11.25\), i.e. 112.5 m. Beyond \(x=15\) the gradient is negative, so the path descends. This kind of problem combines integration with the stationary-point ideas of differentiation.

Worked example 17: Area between a curve and a horizontal line

Find the area of the region enclosed between the curve \(y=x^2-6x+10\) and the line \(y=5\).

Intersections: \(x^2-6x+10=5\Rightarrow x^2-6x+5=0\Rightarrow(x-1)(x-5)=0\), so \(x=1\) and \(x=5\). Between these values the line is above the curve (test \(x=3\): curve \(1\), line \(5\)). Area \(=\int_1^5\left[5-(x^2-6x+10)\right]dx=\int_1^5(-x^2+6x-5)dx=\left[-\dfrac{x^3}3+3x^2-5x\right]_1^5=\left(-\dfrac{125}3+75-25\right)-\left(-\dfrac13+3-5\right)=\dfrac{25}3+\dfrac73=\dfrac{32}3\). The shortcut for a parabola with roots \(\alpha,\beta\) gives \(\dfrac{(\beta-\alpha)^3}6=\dfrac{64}6=\dfrac{32}3\) ✓.

Worked example 18: A definite integral that gives a negative value

Explain why \(\displaystyle\int_0^3(x^2-4)\,dx\) is negative and calculate the area bounded by \(y=x^2-4\), the axes and \(x=3\) in two parts.

The curve crosses the axis at \(x=2\), so part of the region lies below it. \(\int_0^2(x^2-4)dx=\left[\dfrac{x^3}3-4x\right]_0^2=\dfrac83-8=-\dfrac{16}3\): this region is below the axis. \(\int_2^3(x^2-4)dx=\left(9-12\right)-\left(\dfrac83-8\right)=-3+\dfrac{16}3=\dfrac73\): above the axis. The whole integral \(\int_0^3=-\dfrac{16}3+\dfrac73=-3\) is negative because the larger area is below. The geometric area is \(\dfrac{16}3+\dfrac73=\dfrac{23}3\approx7.67\).