Question

Please help, I've been trying to figure this out for half an hour? Integral substitution

Answer 1

\[\int\limits_{1}^{2}\frac{ x^2 }{ \sqrt{2x+1} }\]

Answer 2

Whatever I do, I am not getting the right answer!!

Answer 3

I set \[u=2x+1\] and got \[\frac{ du }{ dx }=2\]
and then \[\frac{ 1 }{ 2 }du=dx\]

Answer 4

Can someone help guide me through the next few steps? I can't figure out what I'm doing wrong

Answer 5

\[x=\frac{u-1}{2}\]

Answer 6

\[x^2=\left(\frac{u-1}{2}\right)^2=\frac{1}{4}(u^2-2u+1)\]

Answer 7

How did you get u-1?

Answer 8

solve \(u=2x+1\) for x

Answer 9

Is this correct: \[\frac{ 1 }{ 2 } \ \int\limits_{1}^{2} \frac{ u^2-2u+1 }{ 4 } \times \frac{ u^{^{1/2}} }{ \frac{ 1 }{ 2 } }\]

Answer 10

why is \(u^{1/2}\) in the numerator?

Answer 11

Oops. I forgot to take the integral sign away. Does it make sense if I take the sign away? And then would I just evaluate it at x=2 and x-1?

Answer 12

=*

Answer 13

you need to integrate first

Answer 14

the entire integrand

Answer 15

I did \[\int\limits_{}^{}u ^{-1/2} = \frac{ u^{1/2} }{ \frac{ 1 }{ 2 } }\]

Answer 16

you can't just integrate that part

Answer 17

you have to integrate the entire thing

Answer 18

So I would just integrate everything after the integrate I set up right?

Answer 19

distribute the \(u^{-1/2}\) then integrate

Answer 20

integrand sign*

Answer 21

also you need to change your limits of integration

Answer 22

Why would I change those?

Answer 23

1 to 2 is for x

you are integrating over u

Answer 24

So how would I figure out what my new limits of integration are?

Answer 25

use u=2x+1

Answer 26

when x=1 u=2(1)+1=3
when x=2 u=2(2)+1=5

Answer 27

so if 1<x<2 then 3<u<5

Answer 28

okay thanks