Using the method of substitution, I = [ x²(1 + 2x³¹)³dx = [ f(u)du where u= du = f(u) = Using the above information: [ x²(1 + 2x³)³dx - f(u)du = || da (in terms (in terms

Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter8: Further Techniques And Applications Of Integration
Section8.2: Integration By Parts
Problem 41E
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Using the method of substitution,
I
! = [ z²(1 + 2x³)³dx = [ f(u)du
where
U=
du: =
f(u) =
Using the above information:
[ x²(1 + 2x³)²dx
= [ f(u)du
=
||
dx
(in terms of u)
(in terms of x)
Note: Don't forget the constant of integration in the last two blanks.
Transcribed Image Text:Using the method of substitution, I ! = [ z²(1 + 2x³)³dx = [ f(u)du where U= du: = f(u) = Using the above information: [ x²(1 + 2x³)²dx = [ f(u)du = || dx (in terms of u) (in terms of x) Note: Don't forget the constant of integration in the last two blanks.
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Follow-up Question

These are incorrect

 

u= 1+2x^3
-lu = 6x^2
f(u) = 4^2/6
Using the above information:
[ x² (1+2x³)²dx
= [ f(u)du
4^3/10 + C
(1+2x^3)^3/10 +C
dx
(in terms of u)
(in terms of x)
Transcribed Image Text:u= 1+2x^3 -lu = 6x^2 f(u) = 4^2/6 Using the above information: [ x² (1+2x³)²dx = [ f(u)du 4^3/10 + C (1+2x^3)^3/10 +C dx (in terms of u) (in terms of x)
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