1. Can someone please explain this integral;

The integral from 0 to 1 of: 2t(sqrt(1 + 2t^2))dt

I tried using substitution

let u=sqrt(1 + 2t^2)

but I keep getting a different answer than the book...

2. Originally Posted by s7b
Can someone please explain this integral;

The integral from 0 to 1 of: 2t(sqrt(1 + 2t^2))dt

I tried using substitution

let u=sqrt(1 + 2t^2)

but I keep getting a different answer than the book...
Substitute $\displaystyle u = 1 + 2 t^2$.

3. ## solved

I like fantastic's method, though since I'm new to the posts, I thought I'd get some practice. So, this is the longer version:
I used 2 substitutions (though you only need one being $\displaystyle u=2+2t^2$ ), being $\displaystyle u=t^2$, du=2tdt and p=1+2u dp=2du so, it should look like this:
$\displaystyle \int_0^1 2t\sqrt{1+2t^2} dt$
$\displaystyle \int_0^1 \sqrt{1+2u} du$ *first substitution*
$\displaystyle (1/2) \int_1^3 \sqrt{p} dp$ *second substitution*
$\displaystyle (1/2) \times (2/3) p^{3/2} |_1^3$
$\displaystyle (1/3) \times (3^{3/2} - 1)$
$\displaystyle \sqrt{3} - (1/3)$

4. oops, my bad, meant to say that the method with 1 sub would be u=1+2t^2, not u=2+2t^2

5. Originally Posted by s7b
Can someone please explain this integral;

The integral from 0 to 1 of: 2t(sqrt(1 + 2t^2))dt

I tried using substitution

let u=sqrt(1 + 2t^2)

but I keep getting a different answer than the book...
just put t^2=u....you will get answer:sqrt(3)-1/3

6. Originally Posted by Mathventure
just put t^2=u....you will get answer:sqrt(3)-1/3
The substitution suggested in post #2 leads to a simpler calculation.