Calculus definite integral help
http://puu.sh/8cflR.jpg
I am not sure how to use substitution, can someone solve this?

Question

Calculus definite integral help
http://puu.sh/8cflR.jpg
I am not sure how to use substitution, can someone solve this?

ganeshie8 · Answer

you cant break the radical,
so whenever u see a radical the first thought should be : to substitute the stuff inside radical as $\large   u$

ganeshie8 · Answer

$\large u =  1-x^2$

ganeshie8 · Answer

Next, differentiate both sides and find out $\large dx$ in terms of $\large du$

anonymous · Answer

how do I do that?

ganeshie8 · Answer

$\large u  =   1-x^2$

*differentiate both sides

$\large  du =  -2x dx$


$\large  \dfrac{-du}{2} = xdx$

ganeshie8 · Answer

After substitution, the integrand becomes :

$\large  \int \sqrt{u}   \dfrac{-du}{2}$

which is same as :

$\large  \dfrac{-1}{2}\int u^{\frac{1}{2}}   du$

ganeshie8 · Answer

fine, so far ?

anonymous · Answer

I think so

ganeshie8 · Answer

next, u may change the bounds according to the substitution and evaluate the integral

ganeshie8 · Answer

knw how to change the bounds ?

anonymous · Answer

I don't think so

ganeshie8 · Answer

its easy :

look at the bounds in given integral,
lower bound =  0
upper bound = 1

ganeshie8 · Answer

and your substitution is :

$\large  u = 1-x^2$

when x = 0,    $u =  ?$
when x = 1,    $u =  ?$

anonymous · Answer

1 and 0

ganeshie8 · Answer

Yes, so the bounds just got flipped.

Then final integral with bounds is :

$\large  \dfrac{-1}{2}\int \limits_1^0 u^{\frac{1}{2}}   du $

ganeshie8 · Answer

we're done.
just evavluate

ganeshie8 · Answer

use below to evaluate :

$\large \int x^n dx = \frac{x^{n+1}}{n+1} + c$

anonymous · Answer

how do I use that equation to evaluate?

ganeshie8 · Answer

$\large  \dfrac{-1}{2}\int \limits_1^0 u^{\frac{1}{2}}   du $

appeal to the above formula :

$\large  \dfrac{-1}{2}  \left(\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}  \Bigg|_1^0  \right)$

ganeshie8 · Answer

simplify and take bounds

ganeshie8 · Answer

dont wry about $\large c$, it goes away when u subtract...

ganeshie8 · Answer

$\large  \dfrac{-1}{2}  \left(\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}  \Bigg|_1^0  \right)$


$\large  \dfrac{-1}{2}  \left(\frac{u^{\frac{3}{2}}}{\frac{3}{2}}  \Bigg|_1^0  \right)$



$\large  \dfrac{-1}{3}  \left(u^{\frac{3}{2}}  \Bigg|_1^0  \right)$

ganeshie8 · Answer

take the bounds now

anonymous · Answer

for 0 it is -1/3 and for 1 it is 0?

ganeshie8 · Answer

$\large  \dfrac{-1}{3}  \left(u^{\frac{3}{2}}  \Bigg|_1^0  \right)$


taking the bounds :

$\large  \dfrac{-1}{3}  \left((0)^{\frac{3}{2}} - (1)^{\frac{3}{2}}     \right)$

ganeshie8 · Answer

simplify

anonymous · Answer

it would be 1/3

anonymous · Answer

Now I'm not entirely sure on how we did this, could we do another?

ganeshie8 · Answer

sure :) u have an example problem ?