Applications of Integration

In this chapter, we present two applications of the definite integral: finding the area between curves in the plane and finding the volume of the 3D objected obtained by rotating about some given axis the area between curves.

Areas between curves

Consider finding the area between two given curves, say

y = f (x)

and

y = g (x)

, over an interval

a \leq x \leq b

figures/area-between-curves-0.svg — Area between two curves

We will first estimate the area and then invoke a limiting process to argue that in the limit our approximations converge to the true area. To start, subdivide the interval

[a, b]

into

n

equal subintervals of length

Δ x = \frac{b - a}{n}

. The subdivision process produces the intervals

[x_{0}, x_{1}]

[x_{1}, x_{2}]

\dots

[x_{n - 1}, x_{n}]

. Consider one of these subinterval, say

[x_{k - 1}, x_{k}]

, and approximate on this subinterval the area between the curves with a rectangle of area

R_{k} = Area of k th rectangle = \underset{height}{\underset{⏟}{(f (c_{k}) - g (c_{k}))}} \underset{width}{\underset{⏟}{Δ x}}

where

c_{k}

is a point chosen inside the interval

[x_{k - 1}, x_{k}]

. The total area between the curves can be approximated by adding the areas of all the rectangles (see Figure 1.2):

Exact Area \approx Area of n rectangles = \sum_{k = 1}^{n} (f (c_{k}) - g (c_{k})) Δ x .

The expression below is called a Riemann sum:

\sum_{k = 1}^{n} (f (c_{k}) - g (c_{k})) Δ x = (f (c_{1}) - g (c_{1})) Δ x + \dots + (f (c_{n}) - g (c_{n})) Δ x .

figures/area-between-curves.svg — Using rectangles to approximate areas between curves; in this case we subdivide the interval $[a, b]$ into $n = 8$ equal subintervals and over each subinterval we approximate the area between the curves with the area of a rectangle.

If now we increase the value of

n

and repeat the process we obtain a new approximation which is closer to the true area between the curves. Hence, to compute exact area we take the limit of the Riemann sum expression as

n \to \infty

. As you may recall from Calculus I, the limit of a Riemann sum is the definition of the integral:

Exact Area = lim_{n \to \infty} \sum_{k = 1}^{n} (f (c_{k}) - g (c_{k})) Δ x = \int_{a}^{b} (f (x) - g (x)) d x .

It is important to emphasize that on the interval of integration

[a, b]

, we need

f (x) \geq g (x) .

In summary, the exact area

A

can be computed as the integral

A = \int_{a}^{b} [\underset{top}{\underset{⏟}{f (x)}} - \underset{bottom}{\underset{⏟}{g (x)}}] d x .

Find the area of the region bounded by the curves

y = e^{x}

y = x

x = 0

, and

x = 1

The curves and the region are shown below

figures/area-between-curves-1.svg — Curves for Example 1.1.1

The area is then

\begin{aligned} A & = \int_{0}^{1} (e^{x} - x) d x \\ = (e^{x} - x^{2}) / 2) |_{0}^{1} \\ = e - \frac{3}{2} \end{aligned}

Find the area of the region bounded by the curves

y = x^{2}

and

y = 2 x - x^{2}

The curves and region is shown below. The curves intersect when

or when

and

. The area is then

figures/area-between-curves-2.svg — Curves for Example 1.1.2

Find the area of the region bounded by the curves

and

The curves and region is shown below. The points of intersection are found by setting

and this gives

The

-values where the curves intersect are therefore

, and

. We must do two integrals to compute the area.

figures/area-between-curves-3.svg — Curves for Example 1.1.3

Find the area of the region bounded by the curves

, and

The curves and region are shown below. If we integrate with respect to

, we will need to do two integrals. On the other hand, the region is bounded on the right by the curve

and on the left by the curve

. The curves intersect at

and

. Integrating with respect to

figures/area-between-curves-4.svg — Curves for Example 1.1.4

Find the area of the region bounded by the curves

, and

The region and curves is shown below. To find where the curves intersect, set

which yields the quadratic

whose roots are

and

. Only interested in

intercept. Integrating with respect to

yields

If instead integrating with respect to

need two integrals:

figures/area-between-curves-5.svg — Curves for Example 1.1.5

Find the area of the region bounded by

and

over the interval

The region and curves is shown below. On the interval

, sine and cosine intersect at

. Need two integrals:

figures/area-between-curves-6.svg — Curves for Example 1.1.6

Find the area of the region bounded by

, and

The region and curves is shown below. Best handled with integration with respect to

figures/area-between-curves-7.svg — Curves for Example 1.1.7

Volumes

Before beginning this subject, see the Maple file suface-revolution.mw found in the directory

Teaching-Resources/Calculus-Resources/Maple

In this section, we will compute volumes of solids using the integral. First recall that the volume of a cylinder is

, see Figure 1.10.

figures/disc.svg — Volume of a cylinder of radius and height is

Consider a function

on the interval

and rotate around the

-axis the region under the graph of

, between

and

, as shown below:

We can estimate the volume of the solid using discs as follows. We subdivide the interval

into

equal subintervals of length

. This produces grid points along the interval

On the

th generic interval

, we choose the mid-point

and consider the disc of radius

and width

, whose volume is therefore

We repeat this process at every interval and obtain

discs whose volumes are added to obtain an approximation to the true volume of the region, such a sum is called a Riemann sum:

increases and we use more discs with smaller width, we obtain a sequence of values that converge to the true volume of the solid:

The limit of a Riemann sum is an integral and thus the volume of the solid is

Show that the volume of a sphere of radius

A sphere can be obtained by rotating about the

-axis a semi-circle of radius

. The function whose graph is the upper semi-circle is

. Therefore, the volume of the sphere is

Find the volume of the solid obtained by rotating about the

-axis the region under the curve

from

Below is the solid and a generic disc. The volume is

Find the volume of the solid obtained by rotating about the

-axis the region bounded by

, and

In this case, we need to use horizontal discs. At a value of

between

, a typical disc has radius

. The volume is therefore

Find the volume of the solid obtained by rotating about the

-axis the region bounded by the curves

and

The solid is shown below. In this case, instead of discs we must use washers. The inner radius is

and the outer radius is

. The volume of a typical washer is having width

is therefore

Therefore, the volume of the solid is

Find the volume of the solid obtained by rotating about the

-axis the region bounded by the curves

, and

At a

value between

, a typical washer has inner radius

and an outer radius

. The volume is therefore

Find the volume of the solid obtained by rotating about the line

the region bounded by the curves

, and

The inner radius of a generic washer is

and outer radius is

. The volume is therefore

Find the volume of the solid obtained by rotating about the line

the region bounded by the curves

and

The inner radius of a generic washer is

and outer radius is

. The volume is therefore

Find the volume of the solid obtained by rotating about the line

the region bounded by the curves

and

The inner radius of a generic washer is

and outer radius is

. The volume is therefore

Find the volume of the solid obtained by rotating about the line

the region bounded by the curves

and

The inner radius of a generic washer is

and outer radius is

. The volume is therefore

Arc Length

Before we begin, we recall the Mean Value Theorem.

is continuous on the interval

and differentiable on

then there is some number

such that

Consider finding the length of a curve that is the graph of a function

over the interval

. Finding the length of a line segment is straightforward and thus we will first use line segments to approximate the graph of

and then use the sum of the lengths of the line segments to approximate the true length of the curve. This process will result in a Riemann sum which will then lead us to an integral for computing the length of the curve. Hence, to begin first subdivide the interval

into

equal subintervals of length

. On each subinterval, use a line segment to approximate

as shown below:

On each subinterval, the length of the line segment can be found by using the Pythagorean theorem:

where

Applying the Mean Value theorem to

on the interval

, there is some point

such that

Therefore, the length of the line segment on the interval

Adding up the lengths of each line segment gives an approximation of the arc length

and taking the limit of the Riemann sum produces the exact value of

By definition, the integral is the limit of a Riemann sum and therefore

Let

. Find the arc length of the graph of

over the interval

The graph of the function is shown below:

Here

and the arc length is then

Find the arc length of the graph of the function

on the interval

The graph of the function is shown below:

Here

and the arc length is therefore

Average Value of a Function

The average value of a given set of numbers

What about the average value of a function

over an interval

? To get an approximation to the average value of

, we could subdivide the interval

into

equally wide subintervals

, evaluate

at the end-points of each subinterval,

, and find the average of the resulting numbers:

The length of each subinterval is

and thus

. Thus,

we obtain the average value of

, which yields the formula

On April 7, 2003 in Geneseo, the temperature (in degrees Fahrenheit)

hours after 9 am was modeled by the function

. Find the average temperature during the period from 9 am to 9 pm on April 7, 2003.

The time

corresponds to 9 am, and thus 9 pm corresponds to

. The average temperature was

Calculus II

1. Applications of Integration

Areas between curves

Volumes

Arc Length

Average Value of a Function