5.3Integration by Substitution

Then, by Leibniz Rule, we have

F 0(x) = f(g(x))g0(x),

and

G0(x) = f(g(x))g0(x)

for all x on [a, b].

It follows that there exists some constant C such that

F (x) = G(x) + C for all x on [a, b]. For x = a we get

0 = F (a) = G(a) + C = 0 + C

and, hence,

C = 0.

Therefore, F (x) = G(x) for all x on [a, b], and hence

Z b

f(g(x))g0(x)dx = F (b)

a

= G(b)

Z g(b)

= f(u)du.

g(a)

This completes the proof of this theorem.

Remark 18 We say that we have changed the variable from x to u through the substitution u = g(x).

Example 5.3.1

where u = x2, du = 2x dx, 3x dx = 32 du.

where u = x2, du = 2x dx, x dx = 12 dx.

Definition 5.3.1 Suppose that f and g are continuous on [a, b]. Then the area bounded by the curves y = f(x), y = g(x), y = a and x = b is defined to be A, where

Z b

A = |f(x) − g(x)| dx.

a

If f(x) ≥ g(x) for all x in [a, b], then

Z b

A = (f(x) − g(x)) dx.

a

If g(x) ≥ f(x) for all x in [a, b], then

Z b

A = (g(x) − f(x)) dx.

a

Example 5.3.2 Find the area, A, bounded by the curves y = sin x, y = cos x, x = 0 and x = π.

graph

Example 5.3.3 Find the area, A, bounded by y = x2, y = x3, x = 0 and x = 2.

graph

We note that x3 ≤ x2 on [0, 1] and x3 ≥ x2 on [1, 2]. Therefore, by definition,

Z 1 Z 2

A = (x2 − x3) dx + (x3 − x2) dx

0 1

=121 + 34 + 121

=32.

Example 5.3.4 Find the area bounded by y = x3 and y = x. To find the interval over which the area is bounded by these curves, we find the points of intersection.

graph

x3 = x ↔ x3 − x = 0 ↔ x(x2 − 1) = 0

↔ x = 0, x = 1, x = −1.

The curve y = x is below y = x3 on [−1, 0] and the curve y = x3 is below the curve y = x on [0, 1]. The required area is A, where

Exercises 5.3 Find the area bounded by the given curves.

Evaluate each of the following integrals: