Question

Integration

Answer 1

\[\int\limits_{0}^{4} \frac{ x^3 }{ \sqrt{x^2 +1} } dx\]

Answer 2

Sending a picture of my work...

Answer 3

put\[\sqrt{x^2+1}=t,x2+1=t^2,2x~dx=2t~dt,x~dx=t~dt\]
when x=0,t=1
when x=4,\[t=\sqrt{17}\]
\[L=\int\limits_{0}^{4}\frac{ x^2*x~dx }{ \sqrt{x^2+1} }\]
i hope now you can solve.

Answer 4

I cant actually sub sqrtx^2+1 in for u

Answer 5

you are also correct..
\[\frac{ 1 }{ 2 }\int\limits_{0}^{17}\left\{ u ^{\frac{ 1 }{ 2 }}-u ^{\frac{ -1 }{ 2 }} \right\}du\]

Answer 6

correction write 1 in place of 0 in the limit

Answer 7

Is that a legal move? Subbing in u for the denominator?

Answer 8

it is simple algebra.

Answer 9

I see. I confused my self. I got it now. Thanks.

Answer 10

yw