Question

Evaluate the integral

∫_0^1{36dx/((2x+1)^3)}

Answer 1

\[\int\limits_{0}^{1}{\frac{36dx}{(2x+1)^3}}\]

Answer 2

Okay:
\[\int\limits_0^1 \frac{36dx}{(2x+1)^3}\]
The first thing you want to do is make a usubstitution.
Let u=2x+1
then du=2dx
or dx=(1/2)du
Re-evaluate your limits plugging in zero you get 1 and plugging in 1 your get 3.
So you have:
\[\frac{36}{2}\int\limits_1^3\frac{du}{u^3}\]
Rewrite it as:
\[18 \int\limits_1^3 u^{-3}du=18[\frac{-1}{2}u^{-2}]_1^3=-9(\frac{1}{3^2}-\frac{1}{1^2})=-9*\frac{-8}{9}=8\]

Answer 3

ok so i know my u=2x+1 du=2 18du=36
18∫u^3du

18u^4/4
(18(2x+1)^4/4)
18(2(1)+1)^4/4-18(2(0)+1)^4/4
364.5-4.5

Answer 4

It becomes u^-3. Not u^3 :P

Answer 5

ok i see where i went wrong thank yu malevolence

Answer 6

No problem :)