Question

problem 14. (1 point)
let ( f ) be a differentiable function.
suppose that ( f(3) = 8, f(9) = 5, f(3) = 7 ) and ( f(9) = 6 ).
further suppose that ( int_{3}^{9} x^2 f(x) dx = 9 ).
then the integral ( int_{3}^{9} x f(x) dx ) is equal to (square).

Explanation:

Step1: Integrate by parts for \(\int_{3}^{9} x^2 f'(x) dx\)

Let \(u = x^2\), \(dv = f'(x)dx\). Then \(du = 2xdx\), \(v = f(x)\). By integration by parts formula \(\int_{a}^{b} u dv=uv|_{a}^{b}-\int_{a}^{b} v du\), we have:
\(\int_{3}^{9} x^2 f'(x) dx = [x^2 f(x)]_{3}^{9} - \int_{3}^{9} 2x f(x) dx\)

Step2: Calculate \([x^2 f(x)]_{3}^{9}\)

\([x^2 f(x)]_{3}^{9}=9^2 f(9)-3^2 f(3)=81\times5 - 9\times8 = 405 - 72 = 333\)

Step3: Substitute into the integration by parts equation

We know \(\int_{3}^{9} x^2 f'(x) dx = 9\), so \(9 = 333 - 2\int_{3}^{9} x f(x) dx\)

Step4: Solve for \(\int_{3}^{9} x f(x) dx\)

Rearrange the equation: \(2\int_{3}^{9} x f(x) dx = 333 - 9 = 324\)
Then \(\int_{3}^{9} x f(x) dx=\frac{324}{2}=162\)

Answer:

\(162\)