Question

Use integration by substitution to find the area under the curve \[y={1 \over sqrt{x}+x}\] between x=1 and x=4.

Answer 1

I've got \[\int_{1}^{4} {1 \over \sqrt{x}+x}\] and that's it. Got no idea for what u should be.

Answer 2

forgot the dx, whatever

Answer 3

Why don't you try letting u=sqrt(x)?

Answer 4

doesn't that mean that du is \[1 \over 2 \sqrt{x}\] then? I couldn't see how that could fit...

Answer 5

Well, that's correct, but you know that u^2=x so that 2u du=dx.

Answer 6

oh yeah... give me a sec to think it through

Answer 7

wait, what about the dx then?

Answer 8

unless we use the du's sqrt{x} for the sqrt{x} in the problem and u^2 for x?

Answer 9

No, the you don't use the du's sqrt(x), but you can substitute u^2 for x.

Answer 10

\[u=\sqrt{x}\]
\[du=\frac{dx}{2\sqrt{x}}\]
\[2\sqrt{x}du=dx\]
\[2udu=dx\]
\[2\int\frac{udu}{u^2+u}\]

Answer 11

ahh.. I see what you did there. thanks.

Answer 12

might as well change limits of integration while you are at it and get
\[2\int_1^2\frac{udu}{u^2+u}\]

Answer 13

yw

Answer 14

ok got it :)