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Mathematics 20 Online
OpenStudy (anonymous):

integrate from 1to 3 for xH square root Hx^2 -1LL

OpenStudy (anonymous):

could you please show me the steps of how they got this?

OpenStudy (anonymous):

Let x^2-1=t differentiate both sided you get 2xdx=dt now xdx=dt/2 so the integral is sqrt(t)/2 which you can integrate as (t^(3/2))/3now t=x^2-1 so the answer is ((x^2-1)^(3/2))/3

OpenStudy (anonymous):

now put limits.

OpenStudy (anonymous):

((x^2-1)^(3/2))/3 <--- shouldn't it be ((x^2-1)^(3/2))/(3/2)

OpenStudy (anonymous):

Yeah but there is sqrt(t)/2 there is two in the denominator

OpenStudy (anonymous):

huh?

OpenStudy (anonymous):

x dx=dt/2 who will take that 2?

OpenStudy (anonymous):

i still don't understand

OpenStudy (anonymous):

When x^2 -1 =t differnentiation results in 2x dx there is a 2 there no

OpenStudy (anonymous):

sorry but i still dont get ya

OpenStudy (mimi_x3):

Let \[u = x^2-1\] \[\int\limits x\sqrt{u}*\frac{du}{2x} \] \[\frac{1}{2} \int\limits\sqrt{u} du\] Then integrate it.

OpenStudy (mimi_x3):

Can you integrate it?

OpenStudy (anonymous):

what ans am i suppoesed to get?

OpenStudy (anonymous):

I don't see an answer on that page anyway the answer is (8^(3/2))/3 16sqrt(2)/3

OpenStudy (mimi_x3):

\[\frac{1}{2} *\frac{u^{\frac{3}{2}}}{3/2} + C \] \[\frac{1}{3} \times u^\frac{3}{2} + C \]

OpenStudy (anonymous):

OOOOOHHHHHHH I GET IT HAHAHAHAHAHAH!

OpenStudy (mimi_x3):

\[\frac{1}{3} * (x^2-1)^\frac{3}{2} + C \] Is this what you got when you integrate it ?

OpenStudy (anonymous):

THANK YOU I FINALLY GET IT HAHAH

OpenStudy (mimi_x3):

np.

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