integrate from 1to 3 for xH square root Hx^2 -1LL
could you please show me the steps of how they got this?
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
now put limits.
((x^2-1)^(3/2))/3 <--- shouldn't it be ((x^2-1)^(3/2))/(3/2)
Yeah but there is sqrt(t)/2 there is two in the denominator
huh?
x dx=dt/2 who will take that 2?
i still don't understand
When x^2 -1 =t differnentiation results in 2x dx there is a 2 there no
sorry but i still dont get ya
Let \[u = x^2-1\] \[\int\limits x\sqrt{u}*\frac{du}{2x} \] \[\frac{1}{2} \int\limits\sqrt{u} du\] Then integrate it.
Can you integrate it?
what ans am i suppoesed to get?
because i'm not getting the same as http://www.wolframalpha.com/input/pdfGet.jsp?id=MSP198219ig4eag86h7i77c00003105gb3d59d57hff&s=8&i=integrate+from+1to+3+for+x(+square+root+(x^2+-1))
I don't see an answer on that page anyway the answer is (8^(3/2))/3 16sqrt(2)/3
\[\frac{1}{2} *\frac{u^{\frac{3}{2}}}{3/2} + C \] \[\frac{1}{3} \times u^\frac{3}{2} + C \]
because i'm not getting the same as http://www.wolframalpha.com/input/pdfGet.jsp?id=MSP198219ig4eag86h7i77c00003105gb3d59d57hff&s=8&i=integrate+from+1to+3+for+x(+square+root+(x^2+-1))
OOOOOHHHHHHH I GET IT HAHAHAHAHAHAH!
\[\frac{1}{3} * (x^2-1)^\frac{3}{2} + C \] Is this what you got when you integrate it ?
because i'm not getting the same as http://www.wolframalpha.com/input/pdfGet.jsp?id=MSP198219ig4eag86h7i77c00003105gb3d59d57hff&s=8&i=integrate+from+1to+3+for+x(+square+root+(x^2+-1))
THANK YOU I FINALLY GET IT HAHAH
np.
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