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] =
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…
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…