Integration in parts to find area between 16ln(x), xln(x)

[insanerp]

Find the area of the region bounded by the curves y = 16ln(x) and y = xln(x)

I know area ?r^2 so ?[(16ln(x))^2-(xln(x))^2]

I separated the terms and I am able to integrate the first half, but the second half where u=(xln(x))^2, v'=1, v=x, and I am having difficulty solving for u' and what are my limits to solve for area?
Thanks

 

[stapel]

Find the area of the region bounded by the curves y = 16ln(x) and y = xln(x)

I know area ?r^2 so ?[(16ln(x))^2-(xln(x))^2]

Normally, the area between two curves is found by subtracting the "lower" curve from the "higher", and integrating (summing) the resulting expression (representing "slices" of height). On what basis did you conclude that the area would be a circle, or somehow related to circles...?

Please be complete. Thank you!

Eliz.

 

[skeeter]

16ln(x) = xln(x)
16ln(x) - xln(x) = 0
ln(x)[16 - x] = 0
x = 1, x = 16

\(\displaystyle V = \int_1^{16} (16-x)\ln(x) \, dx\)

now do the integration by parts using \(\displaystyle u = \ln(x)\) and \(\displaystyle dv = (16-x)\)