integration problem

[tandoorichicken]

Hello everybody
I'm working on a section for integration by substitution and I came across an integral that I don't know how to do a substitution for

\int_{1}^{2} x\sqrt{x-1} \,dx

How can I do this problem?

[vsage]

This isn't simple substitution. You can multiply the equation fancily by 1 and then use substitution. Tell me if this is too vague.

Edit: It seems someone else found it just by substitution. Oh well, I like my way better :p

[Clausius2]

Hello everybody
I'm working on a section for integration by substitution and I came across an integral that I don't know how to do a substitution for

\int_{1}^{2} x\sqrt{x-1} \,dx

How can I do this problem?

t=\sqrt{x-1}
x=t^2+1
dx=2tdt

Your integral is:

\int_{make it you}^{make it you} (t^2+1)t^2 2dt

[arildno]

Another choice is:
\int_{1}^{2}x\sqrt{x-1}dx=\int_{1}^{2}(x-1)^{\frac{3}{2}}dx+\int_{1}^{2}\sqrt{x-1}dx

[tandoorichicken]

Thanks, everybody. Would you all care to check my work please?

\int_{1}^{2} x\sqrt{x-1} \,dx
u = \sqrt{x-1}
x = u^2 + 1
\,du = \frac{1}{2\sqrt{x-1}}\,dx
\,dx = 2\sqrt{x-1} \,du = 2u\,du
\int_{1}^{2} x\sqrt{x-1} \,dx = \int_{0}^{1} (u^2 + 1)u 2u\,du = \int_{0}^{1} 2u^4 + 2u^2 \,du = 2\int_{0}^{1} u^4 \,du + 2\int_{0}^{1} u^2 \,du = 2 + 2 = 4

Is this correct?

[arildno]

Do the following integrals once more:
\int_{0}^{1}u^{4}du,\int_{0}^{1}u^{2}du

[tandoorichicken]

Oh, whoops. Thanks for pointing that out.

\int u^4 \,du = \frac{u^5}{5} , \int u^2 \,du = \frac{u^3}{3}
So then it becomes
\frac{2}{5} + \frac{2}{3} = \frac{16}{15}

[arildno]

Seems much better! :biggrin: