contestada

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).

Respuesta :

LammettHash

Integrating by parts, we have

[tex]\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[/tex]

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

[tex]15=-u(1)v(1)-20\implies\boxed{u(1)v(1)=-35}[/tex]

Otras preguntas