Question

Find the volume of the solid between the curve f(x)=sqrtx(lnx) and the x-axis and vertical lines x=1 and x=2.

Answer 1

You must mean the area ... or is the region being revolved about some axis?

Answer 2

In any case, is it \(f(x)=\sqrt x\cdot\ln x\) or \(f(x)=\sqrt{ x\ln x}\), or some other variation?

Answer 3

\[\sqrt{x}*lnx\]
I may have solved this since posting, but I'd be curious to see what others came up with.

Answer 4

Assuming it's area ...
\[\int_1^2\sqrt x~\ln x~dx\]
Integration by parts seems to be the best way:
\[\begin{matrix}u=\ln x&&&dv=\sqrt x~dx\\
du=\frac{dx}{x}&&&v=\frac{2}{3}x^{3/2}\end{matrix}\]
So the integral is equivalent to
\[\left[\frac{2}{3}x^{3/2}\ln x\right]_1^2-\frac{2}{3}\int_1^2\sqrt x~dx\]

Answer 5

I'm wondering if it could also be done with a substitution like \(x=e^u\), so \(dx=e^u~du\), and the integral becomes
\[\large \int_0^{e^2} \sqrt{e^u}~(\ln e^u)~e^u~du=\int_0^{e^2}ue^{3/2~u}~du\]
Seems that way! Integration by parts from here.