Evaluate the integral by making the given substitution. xV+ 14 dx + 14 dx, u = x3 + 14

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Step 2
If u = x3 + 14 is substituted into
xV + 14 dx, then
have
(u)/2 dx =
u1/2 x2 dx-
we
We must also convert x dx into an expression involving u.
We know that du =
3x2 dx, and so x² dx =
du.
Step 3
Now, if u = x + 14, then
X3+14 dx =
u/2 du-
%3D
This evaluates as
14) (),
2u/2
ul/2 du =
9.
+ C
Enter your answer in terms of u. + C.
Step 4
Since u = x + 14, then converting back to an expression in x we get
+ C =
9
13
1/3
Transcribed Image Text:Step 2 If u = x3 + 14 is substituted into xV + 14 dx, then have (u)/2 dx = u1/2 x2 dx- we We must also convert x dx into an expression involving u. We know that du = 3x2 dx, and so x² dx = du. Step 3 Now, if u = x + 14, then X3+14 dx = u/2 du- %3D This evaluates as 14) (), 2u/2 ul/2 du = 9. + C Enter your answer in terms of u. + C. Step 4 Since u = x + 14, then converting back to an expression in x we get + C = 9 13 1/3
Evaluate the integral by making the given substitution.
dx, u = x3 + 14
Step 1
We know that if u = f(x), then du = f '(x) dx. Therefore, if u = x + 14, then
3x
du =
dx.
3r2
Step 2
If u = x + 14 is substituted into
x²V x3 + 14 dx, then we have
x² (u)!/2 dx =
u/2 x2 dx-
We must also convert x² dx into an expression involving u.
We know that du = 3x² dx, and so x² dx =
du.
Step 3
Now, if u = x + 14, then
u!/2
u/2 du-
%3D
This evaluates as
2( 3
2u3/2
1/3
Transcribed Image Text:Evaluate the integral by making the given substitution. dx, u = x3 + 14 Step 1 We know that if u = f(x), then du = f '(x) dx. Therefore, if u = x + 14, then 3x du = dx. 3r2 Step 2 If u = x + 14 is substituted into x²V x3 + 14 dx, then we have x² (u)!/2 dx = u/2 x2 dx- We must also convert x² dx into an expression involving u. We know that du = 3x² dx, and so x² dx = du. Step 3 Now, if u = x + 14, then u!/2 u/2 du- %3D This evaluates as 2( 3 2u3/2 1/3
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