Need help with calculus
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Need help with calculus

[From: ] [author: ] [Date: 12-05-02] [Hit: ]
interval [0,dx/[(2 x + 1)^3] =-So if u = 2x + 1,Go ahead and place the (1/2) outside the integral to clean it up. Also, bring u to the numerator, since 1/u^3 = u^-3.......
how do u solve these??

Evaluate the integral by making the given substitution u = 2 x + 1.
dx/[(2 x + 1)^3] =

Evaluate the following definite integral. interval [0,1]
dx/[(2 x + 1)^3] =

-
So if u = 2x + 1, then:

du = 2dx ~~~~> dx = (1/2)du

Plug in u for (2x +1) and (du/2) for dx and get:

∫[((1/2)du)/u^3]

Go ahead and place the (1/2) outside the integral to clean it up. Also, bring u to the numerator, since 1/u^3 = u^-3. You should get:

= (1/2)∫u^(-3) du
= (1/2)(-1/2)[u^(-2)] + C
= (-1/4)u^(-2) + C

Plug u back in:

= -1/[4(2x + 1)^2] + C

To find the definite integral, subtract the integral function at 0 from the integral function at 1

= -1/[4(3)^2] - -1/[4(1)^2]
= -1/(4*9) + 1/(4)
= -1/36 + 1/4
= -1/36 + 9/36
= 8/36
= 2/9

Check: http://www.wolframalpha.com/input/?i=int…
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