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Application of Integrals
Introduction
Area under Simple Curves
Practice: Areas under Curves
Miscellaneous Examples
Chapter Review: Mixed Practice
Summary
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Application of IntegralsArea under Simple Curves

Reading time: ~30 min

In Chapter 7 you evaluated definite integrals as limits of sums and, much faster, with the Fundamental Theorem of Calculus. Now we turn them into a tool for measuring area. Take the region bounded by a curve y = f(x), the x-axis and the ordinates x = a and x = b — the region PQRSP in Fig. 8.1.

Think of this region as a huge number of very thin vertical strips standing side by side. Pick any one strip: its height is y, the value of f at that spot, and its width is a tiny dx. It is very nearly a rectangle, so its area is

dA =

This is the elementary area. The strip can stand anywhere between the two ordinates; its position is fixed by some value of x between a and b.

Fig. 8.1 — the region under the curve y = f(x) from x = a to x = b, and one thin strip.

The area of the whole region is what you get by adding up the elementary areas of all the strips across it. Adding infinitely many infinitely thin pieces is exactly what an integral does:

The area of the region bounded by the curve y = f(x), the x-axis and the ordinates x = a and x = b (b > a) is

A = \int_a^b dA = \int_a^b y\, dx = \int_a^b f(x)\, dx.

You have already used this formula once, on Harpreet's field: there f(x) = x^2, a = and b = .

Vertical or Horizontal?

Strips do not have to stand upright. Some regions are bounded by a curve that is easier to write as x = g(y), together with the y-axis and two horizontal lines y = c and y = d. Then it is natural to slice the region into thin horizontal strips. A horizontal strip has length x and thickness dy, so its elementary area is x\, dy, and

The area of the region bounded by the curve x = g(y), the y-axis and the lines y = c and y = d is

A = \int_c^d x\, dy = \int_c^d g(y)\, dy.

How do you choose? A good strip has both of its ends on the same two boundaries all the way across the region. Then one formula gives its length, and one integral gives the area.

Pick a region, then try a kind of strip. Strips that end on the wrong boundary turn red.

Region 2 is the situation of Fig. 8.2: a curve x = g(y), the y-axis and the lines y = c, y = d.

Did you notice that Regions 1 and 2 are mirror images of each other in the line y = x? The curves y = \frac{x^2}{4} and x = \frac{y^2}{4} are reflections, so the two regions have the same area, and \int_2^4 \frac{x^2}{4}\, dx and \int_2^4 \frac{y^2}{4}\, dy are the same integral with the letter changed. Work one of them out: \left[\frac{x^3}{12}\right]_2^4 = \frac{64 - 8}{12} = .

When the Curve Dips Below the Axis

So far every curve has stayed above the x-axis. Suppose instead that f(x) < 0 all the way from x = a to x = b, as in Fig. 8.3 of your textbook. Then every strip has a negative height y, every y\, dx is negative, and so the integral \int_a^b f(x)\, dx comes out .

But an area is a size, and a size cannot be negative. Only the numerical value counts, so we take the absolute value.

Remark. If the curve lies below the x-axis from x = a to x = b, the area bounded by the curve, the x-axis and these ordinates is \left\lvert \int_a^b f(x)\, dx \right\rvert.

Often a curve is partly below the axis and partly above it, as in Fig. 8.4. The curve below is y = -2\sin\frac{\pi x}{3} with a = 1 and b = 5. Integrating over the teal piece gives a negative number, A_1 < 0; integrating over the purple piece gives A_2 > 0. The area is the sum of the two sizes:

A = \lvert A_1 \rvert + A_2.

Fig. 8.4 — below the x-axis from x = a to the crossing at x = 3, above it from there to x = b.

For this curve A_1 = -\frac{9}{\pi} and A_2 = \frac{9}{\pi}. Integrating straight from a to b gives A_1 + A_2 = , as if there were no region at all. The true area is \lvert A_1 \rvert + A_2 = \frac{18}{\pi}.

Tip

Before you integrate, find every point between the limits where the curve crosses the x-axis. Split the integral there, integrate each piece, and add the absolute values. The Miscellaneous Examples do exactly this.

Example 1: A Circle from One Quarter

Example 1 : Find the area enclosed by the circle x^2 + y^2 = a^2.

A circle is symmetric about both axes, so its area is 4 times the area of the quarter in the first quadrant. Before the algebra, measure a few quarters yourself.

Drag the amber handle along the x-axis to sweep the area under the arc. The readout is the integral so far. Change the radius with the slider.

Radius r = ${r}

Fig. 8.5 — the arc of the circle in the first quadrant, swept by vertical strips.

Solution : From Fig. 8.5, the whole area enclosed by the circle is

4 \times (area of the region AOBA bounded by the arc, the x-axis and the ordinates x = 0 and x = a)

= 4\int_0^a y\, dx (taking vertical strips).

From x^2 + y^2 = a^2 we get y = \pm\sqrt{a^2 - x^2}. The region AOBA lies in the first quadrant, so y is and we take y = \sqrt{a^2 - x^2}.

\text{Area} = 4\int_0^a \sqrt{a^2 - x^2}\, dx

= 4\left[\frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}\right]_0^a (the standard form from Chapter 7)

= 4\left[\left(\frac{a}{2} \times 0 + \frac{a^2}{2}\sin^{-1} 1\right) - 0\right]

Since \sin^{-1} 1 = , the area is 4 \times \frac{a^2}{2} \times \frac{\pi}{2} = .

That is the familiar \pi a^2 — now proved with an integral. Your sweeps agree: a quarter of radius 3 came to \frac{9\pi}{4} \approx 7.07.

Alternatively, use horizontal strips, as in Fig. 8.6 of your textbook. A horizontal strip at height y runs from the y-axis across to the circle, so its length is x = \sqrt{a^2 - y^2}, and

\text{Area} = 4\int_0^a x\, dy = 4\int_0^a \sqrt{a^2 - y^2}\, dy

= 4\left[\frac{y}{2}\sqrt{a^2 - y^2} + \frac{a^2}{2}\sin^{-1}\frac{y}{a}\right]_0^a

= \pi a^2.

Why does it give the same answer? It is the same integral as before with the letter y in place of x — the circle's equation does not change when x and y swap places.

Example 2: Squashing the Circle

Example 2 : Find the area enclosed by the ellipse \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.

An ellipse is a circle that has been squashed (or stretched) in one direction. Compare the ellipse with the circle of radius a: at every x, the ellipse's strip is \frac{b}{a} times as tall as the circle's. Squash every strip by the same factor and the whole area is squashed by that factor too.

Drag a and b to complete the missions.

a = ${a}

b = ${b}

Fig. 8.7 — the ellipse with semi-axes a and b; the shaded quarter is the region AOBA.

Solution : The ellipse is symmetric about both axes, so its area is 4 times the area of the region AOBA in the first quadrant:

\text{Area} = 4\int_0^a y\, dx (taking vertical strips).

From \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 we get y = \pm\frac{b}{a}\sqrt{a^2 - x^2}, and in the first quadrant y is positive. So

\text{Area} = 4\int_0^a \frac{b}{a}\sqrt{a^2 - x^2}\, dx

= \frac{4b}{a}\int_0^a \sqrt{a^2 - x^2}\, dx

The integral that is left is the one from Example 1: the area of a quarter of the circle of radius a, so \int_0^a \sqrt{a^2 - x^2}\, dx = .

Hence \text{Area} = \frac{4b}{a} \times \frac{\pi a^2}{4} = .

Written out in full, the working is

\frac{4b}{a}\left[\frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}\right]_0^a

= \frac{4b}{a} \cdot \frac{a^2}{2} \cdot \frac{\pi}{2} = \pi ab.

Alternatively, use horizontal strips, as in Fig. 8.8 of your textbook. Now x = \frac{a}{b}\sqrt{b^2 - y^2}, and

\text{Area} = 4\int_0^b x\, dy = \frac{4a}{b}\int_0^b \sqrt{b^2 - y^2}\, dy

= \frac{4a}{b} \cdot \frac{\pi b^2}{4} = \pi ab.

Why the same answer? This time we start from the circle of radius b and stretch it sideways by \frac{a}{b}. And when a = b, the ellipse is a circle and \pi ab = \pi a^2: Example 1 is a special case of Example 2.

Exam tip

In a board answer, draw a rough sketch, shade the region, and write "by symmetry, area = 4 × area in the first quadrant" before you integrate. Each of these lines carries marks. Leave \pi in the final answer unless a value is given.

Speed Round: Which Integral?

Every area problem starts with the set-up: the right strips, the right height or length, the right limits. Look at the shaded region and pick the integral that measures it.

Get five right to move on.

Sina