10/9/22, 8:40 PM
5.6 HW - Definite Integral Substitutions-Anthony Mathai
Instructor: Strashimir Popvassilev
Assignment: 5.6 HW - Definite Integral
Course: MATH 212-FG Fall 2022 Calc II
Substitutions
w Intro Multivar Funct 4pm
Student: Anthony Mathai
Date: 10/09/22
Use the Substitution Formula to evaluate the following integrals.
1
22
a.∫ t 9 + 16t dt
b. ∫ t 9 + 16t dt
0
1
a. If g′ is continuous on the interval​[a,b] and f is continuous on the range of​g, then the following Substitution Formula is valid.
b
g(b)
∫ f(g(x)) • g′(x) dx = ∫ f(u) du
a
g(a)
Determine the most appropriate substitution for u = ​g(t).
u = 9 + 16t
Find du.
du = 16 dt
Express t in terms of u.
t=
u−9
16
Express dt in terms of du.
1
dt =
16
du
Recall that u = 9 + 16t. Now transform the limits of integration. When t = ​0, u = ​g(0) = 9.
When t = ​1, u = ​g(1) = 25.
Now rewrite the integral in terms of u. Recall that u = 9 + 16t and du = 16 dt.
25
∫
9
u−9
256
u du
This integrand can be written using fractional exponents as follows.
25
1
u−9 2
∫
u du
256
9
Move the constant denominator outside the integral.
25
1
256
∫ (u − 9)u
1
2
du
9
n
Splitting the integrand into two parts results in integrals that have integrands of the form u which is a basic form that can be
n
integrated using the rule ∫u du =
u
n+1
n+1
.
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∫(u − 9)u
5.6 HW - Definite Integral Substitutions-Anthony Mathai
1
3
2
2
du = ∫u
1
du − 9∫u
2
du
Find the value of each integral.
3
∫u
2
5
1
3
2 2
2 2
2
du = u and ∫u du = u
5
3
The Fundamental Theorem of Calculus states that if f is continuous over​[a,b] and F is any antiderivative of f on​[a,b], then the
definite integral can be found from the following formula.
b
∫ f(x) dx = F(b) − F(a)
a
Use the Fundamental Theorem of Calculus to evaluate these definite integrals. Start with the first one.
25
∫u
3
2
5
du =
9
25
2 2
5764
u
=
5
5
9
Now evaluate the second integral.
25
∫u
1
2
3
du =
9
25
2 2
196
u
=
3
3
9
Finally, compute the desired integral.
25
25
1
u−9 2
1
∫
u du =
256
256
∫u
9
25
3
2
du − 9 ∫ u
9
1
2
353
du =
160
9
1
Thus ∫ t 9 + 16t dt =
0
353
160
.
b. Since the integrand is the same as the one in part​(a), make the same substitution u = 9 + 16t. Find the new limits of
integration.
22
361
1
u−9 2
∫ t 9 + 16t dt = ∫
u du
256
1
25
The indefinite integral will be the same.
361
361
1
u−9 2
1
∫
u du =
256
256
25
=
=
1
256
361
3
∫ u
2
du − 9 ∫ u
25
1
2
du
25
5
361
3
361
2 2
2 2
u
−9
u
5
3
25
25
592,991
160
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10/9/22, 8:40 PM
22
Thus ∫ t 9 + 16t dt =
1
5.6 HW - Definite Integral Substitutions-Anthony Mathai
592,991
160
.
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