H2 Math Applications of Integration: Study Guide

JC · H2 Mathematics · Study guide

Set up split and parametric areas, volumes about coordinate and shifted axes, disc–washer changes, symmetric rotations, Riemann sums and rigorous rectangle bounds.

Concepts · worked methods · graphs10 min read · practice linked below

Draw the strip before writing the integral

For area, identify “upper minus lower” with vertical strips or “right minus left” with horizontal strips. For volume, measure each radius from the axis of rotation. Find every point where a boundary, sign or radius changes before deciding the limits.

Area: a boundary changes inside the region

One region, two valid strip directions

Find the area of the first-quadrant region bounded by $y=x^2$ , its tangent $y=2x-1$ , and the x-axis.

The curve and line meet when $x^2=2x-1$ , giving $(x-1)^2=0$ , so the contact is $(1,1)$ . The line meets the axis at $x=1/2$ .

With vertical strips, the lower boundary is the axis until x = 1/2, then the tangent:

\begin{aligned}A&=\int_0^{1/2}x^2\,dx\\&\quad+\int_{1/2}^{1}[x^2-(2x-1)]\,dx,\\&=\left[\frac{x^3}{3}\right]_0^{1/2}\\&\quad+\left[\frac{(x-1)^3}{3}\right]_{1/2}^{1},\\&=\frac1{24}+\frac1{24}=\frac1{12}.\end{aligned}

Horizontal strips avoid the split. From $y=x^2$ , use the positive branch $x=\sqrt y$ ; from the line, $x=(y+1)/2$ . For $0\leq y\leq1$ , the line is the right boundary:

\begin{aligned}A&=\int_0^1\left(\frac{y+1}{2}-\sqrt y\right)dy,\\&=\left[\frac{y^2}{4}+\frac y2-\frac23y^{3/2}\right]_0^1,\\&=\frac14+\frac12-\frac23=\frac1{12}.\end{aligned}

Matching answers check both setups. The better strip direction is the one that keeps the boundaries simple.

A signed integral is not the total area

Find the total area between $y=x^2-1$ and the x-axis from $x=-2$ to $x=2$ .

The zeros are ±1. The curve lies below the axis between them and above it elsewhere. Use symmetry after splitting the signs:

\begin{aligned}A&=2\int_0^1(1-x^2)\,dx\\&\quad+2\int_1^2(x^2-1)\,dx,\\&=2\left[x-\frac{x^3}{3}\right]_0^1\\&\quad+2\left[\frac{x^3}{3}-x\right]_1^2,\\&=\frac43+\frac83=4.\end{aligned}

The single signed integral is $4/3$ , because the part below the axis subtracts. Its absolute value is not the total area.

Parametric area: replace dx as well as y

Subtract the tangent area with parameter limits

A curve has $x=3t^3$ , $y=3/(1+t^2)$ . The tangent at $t=-1$ is $y=x/6+2$ . Find the area between the curve, this tangent and the y-axis for $-1\leq t\leq0$ .

Here $dx/dt=9t^2\geq0$ , and $x$ runs from −3 to 0. The curve is above the tangent on this interval. The area under the tangent is a trapezium with heights $3/2$ and 2.

\begin{aligned}A&=\int_{-3}^{0}y\,dx-\frac12\left(\frac32+2\right)3,\\&=\int_{-1}^{0}\frac{3}{1+t^2}(9t^2)\,dt-\frac{21}{4},\\&=27\int_{-1}^{0}\left(1-\frac1{1+t^2}\right)dt\\&\quad-\frac{21}{4},\\&=27[t-\tan^{-1}t]_{-1}^{0}-\frac{21}{4},\\&=\frac{87-27\pi}{4}.\end{aligned}

The answer is about 0.5442 square units. At $t=0$ , both parameter derivatives vanish, but this does not spoil this area substitution; the smooth parameter integral is still valid.

Volume of revolution: find the radii

Rotate about the y-axis: use x as a function of y

The region bounded by $x^2-y^2=16$ , $x=5$ , and the x-axis in the first quadrant is rotated completely about the y-axis. Find the volume.

The curve meets $x=5$ at $y=3$ . A horizontal strip runs from $x=\sqrt{y^2+16}$ to $x=5$ , so its rotation produces a washer.

\begin{aligned}R&=5,\qquad r=\sqrt{y^2+16},\\V&=\pi\int_0^3(R^2-r^2)\,dy,\\&=\pi\int_0^3[25-(y^2+16)]\,dy,\\&=\pi\left[9y-\frac{y^3}{3}\right]_0^3,\\&=18\pi\text{ units}^3.\end{aligned}

The washer area is the difference of the squared radii. Squaring the difference of the radii gives a different shape.

Rotate about a shifted horizontal axis

The region between $y=2x$ and $y=x^2$ , $0\leq x\leq2$ , is rotated completely about $y=-1$ . Find the volume.

Both curves are above the axis of rotation. Therefore the upper curve gives the outer radius; each radius includes the extra distance 1.

\begin{aligned}R&=2x+1,\qquad r=x^2+1,\\V&=\pi\int_0^2[(2x+1)^2-(x^2+1)^2]dx,\\&=\pi\int_0^2(-x^4+2x^2+4x)\,dx,\\&=\pi\left[-\frac{x^5}{5}+\frac{2x^3}{3}+2x^2\right]_0^2,\\&=\frac{104\pi}{15}\text{ units}^3.\end{aligned}

A translation of the rotation axis changes both radii. It does not change the intersection limits.

Cut the region where a disc becomes a washer

A region crosses the axis of rotation

The triangular region has vertices $(-1,0)$ , $(2,0)$ and $(2,3)$ . Rotate it completely about the y-axis. Find the volume.

The sloping boundary is $x=y-1$ . At height $y$ , the strip is $y-1\leq x\leq2$ . For $0\leq y\leq1$ , the strip crosses the axis: rotation fills a disc of radius 2. The part on the left creates no extra volume. For $1<y\leq3$ , it creates a washer with inner radius $y-1$ .

\begin{aligned}V&=\pi\int_0^1 2^2\,dy\\&\quad+\pi\int_1^3[2^2-(y-1)^2]dy,\\&=4\pi+\pi\left[4y-\frac{(y-1)^3}{3}\right]_1^3,\\&=4\pi+\pi\left(8-\frac83\right),\\&=\frac{28\pi}{3}\text{ units}^3.\end{aligned}

The switch is at y = 1, where the left boundary reaches the axis. Do not subtract its square while the strip still crosses the axis.

y = 1.0: the strip starts at x = 0.0. Disc radius 2, cross-sectional area 4π. The inner hole begins above y = 1.

A half-turn can already fill a symmetric solid

The region $x^2\leq y\leq4$ is rotated through $\pi$ radians about the y-axis. Find the volume.

At each $y\in[0,4]$ , the region contains the whole segment $-\sqrt y\leq x\leq\sqrt y$ . This diameter sweeps out a full disc in a half-turn: its two halves sweep opposite semicircles.

\begin{aligned}V&=\pi\int_0^4(\sqrt y)^2\,dy,\\&=\pi\left[\frac{y^2}{2}\right]_0^4,\\&=8\pi\text{ units}^3.\end{aligned}

Halving the full-turn integral would undercount this solid. Decide what is actually swept out; overlapping swept parts count only once.

Integration as the limit of a sum

Identify width, height and endpoints

Evaluate $\displaystyle L=\lim_{n\to\infty}\frac1n\sum_{r=1}^n(e^{3r/n}+1)$ . Explain whether the finite sums are above or below the limiting area.

Each rectangle has width $h=1/n$ and right endpoint $x_r=r/n$ , so its height is $f(x_r)$ , where $f(x)=e^{3x}+1$ . The interval is $[0,1]$ .

\begin{aligned}L&=\int_0^1(e^{3x}+1)\,dx,\\&=\left[\frac{e^{3x}}3+x\right]_0^1,\\&=\frac{e^3+2}{3}.\end{aligned}

Since $f'(x)=3e^{3x}>0$ , each right-endpoint rectangle lies above its curve segment. Hence every finite sum is greater than L. If the original expression contains an extra $n$ outside the exponential terms, write it as a sum of $n$ ones first.

Use upper and lower rectangles to prove a bound

For $f(x)=1/x$ on $[1,2]$ , let $S_n$ be the sum of n right-endpoint rectangles. Prove $\displaystyle S_n<\ln2<S_n+\frac1{2n}$ .

\begin{aligned}h&=\frac1n,\qquad x_r=1+\frac rn,\\S_n&=\frac1n\sum_{r=1}^n\frac1{1+r/n},\\&=\sum_{r=1}^n\frac1{n+r}.\end{aligned}

The function is strictly decreasing, so right-endpoint heights give a lower sum. Left-endpoint heights give an upper sum Uₙ.

\begin{aligned}U_n&=\frac1n\sum_{r=0}^{n-1}\frac1{1+r/n},\\U_n-S_n&=\frac1n[f(1)-f(2)],\\&=\frac1n\left(1-\frac12\right),\\&=\frac1{2n}.\end{aligned}

\begin{aligned}S_n&<\int_1^2\frac{dx}{x}<U_n,\\S_n&<\ln2<S_n+\frac1{2n}.\end{aligned}

For n = 2, the lower bound is

\begin{aligned}S_2&=\frac13+\frac14=\frac7{12}.\end{aligned}

For n = 3, it improves to

\begin{aligned}S_3&=\frac14+\frac15+\frac16=\frac{37}{60}.\end{aligned}

Thus $7/12<37/60<\ln2$ . Identify which vertex touches the curve before choosing an upper or lower sum.

Before you finish

Check intersections in the original equations, choose a strip direction, split at changing boundaries, measure radii from the stated axis, and use square or cubic units as appropriate. For a limit of a sum, state the rectangle width, height and interval.

Applications of integration practice questions