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(6/15/09)
Math 20B. Lecture Examples.
Section 5.6. Substitution method†
Theorem 1 (Integration by substitution in indefinite integrals) If y = f (u) is continuous on
an open interval and u = u(x) is a differentiable function that is defined on an interval and
whose values are in the domain of f , then for x in the domain of u,
f (u(x)) u′(x) dx =
f (u) du (1a)
or, in Leibniz notation, ∫
f (u)
du
dx dx^ =
f (u) du. (1b)
Example 1 Find the antiderivative
(x^2 + 1 )^5 (2x) dx.
Answer:
(x^2 + 1)^5 (2x) dx = 16 (x^2 + 1)^6 + C
Example 2 Perform the integration
x^3
x^4 + 16 dx.
Answer:
x^3
√ x^4 + 16 dx = 16 (x^4 + 16)^3 /^2 + C
Example 3 Find the antiderivatives
cos (
√ x)
x
dx.
Answer:
cos(√x) √x dx = 2 sin
x
Example 4 Figure 1 shows the region between y =
10x
(x^2 + 1 )^2
and the x-axis for
0 ≤ x ≤ 3. Find its area. (Make a change of variables in the indefinite
integral.)
1 2 3 x
y
y = 10 x (x^2 + 1)^2
FIGURE 1
Answer:
10 x (x^2 + 1)^2
dx = 5 x^2 + 1
†Lecture notes to accompany Section 5.6 of Calculus, Early Transcendentals by Rogawski
Math 20B. Lecture Examples. (6/15/09) Section 5.6, p. 2
Example 5 Find the value of
0
sin (πx) dx by making a change of variables in the
indefinite integral.
Answer:
sin(πx) dx = − (^) π^1 cos(πx) + C •
0
sin(πx) dx = (^) π^2
In Examples 4 and 5 above, we evaluated definite integrals by making substitutions in the indefinite integrals. We could make the changes of variables in the definite integrals instead by using the next result.
Theorem 2 (Integration by substitution in definite integrals) If y = f (u) is continuous on
an open interval and u = u(x) is a differentiable function that is defined on an interval and
whose values are in the domain of f , then for a and b in the domain of u,
∫ x=b
x=a
f (u(x)) u′(x) dx =
∫ u=u(b)
u=u(a)
f (u) du (2a)
or, in Leibniz notation,
∫ x=b
x=a
f (u)
du
dx dx^ =
∫ u=u(b)
u=u(a)
f (u) du. (2b)
Example 6 Evaluate
0
e−2x^ dx by making a change of variables in the definite integral.
Answer:
0
e−^2 x^ dx = 12 (1 − e−^2 )
Example 7 Evaluate
0
x
1 − x dx by making a change of variables in the definite
integral.
Answer:
∫ x=
x=
x
√ 1 − x dx = 154
Interactive Examples
Work the following Interactive Examples on Shenk’s web page, http//www.math.ucsd.edu/˜ashenk/:‡
Section 6.8: Examples 1–
‡The chapter and section numbers on Shenk’s web site refer to his calculus manuscript and not to the chapters and sections of the textbook for the course.
