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...
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...
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) $