Question

Suppose we know that the functions r and s are are everywhere differentiable and that u(3)=0. Suppose we also know that for 1 ≤ x ≤ 3, the area between the x-axis and the non negative functions h(x)=u(x)dv/dx is 15, and that on the same interval, the area between the x-axis and the non negative function k(x) = v(x)du/dx is 20. Determine u(1)v(1).

Answer 1

Integrating by parts, we have

$\displaystyle\int_1^3\underbrace{u(x)\dfrac{\mathrm dv}{\mathrm dx}}_{h(x)}\,\mathrm dx=u(3)v(3)-u(1)v(1)-\int_1^3\underbrace{v(x)\dfrac{\mathrm du}{\mathrm dx}}_{k(x)}\,\mathrm dx$

We're given $u(3)=0$ and $\int_1^3h(x)\,\mathrm dx=15$ and $\int_1^3k(x)\,\mathrm dx=20$. So we have

$15=-u(1)v(1)-20\implies\boxed{u(1)v(1)=-35}$