Prob 12. Suppose f(x) is differentiable, f(1)=3,f(3)=1 and int _(1)^3xf'(x)dx=13 What is the average value of f on the interval [1,3]

Let the average value of $f(x)$ on the interval $[1,3]$ be denoted by $A$. Then<br />$$A = \frac{1}{3-1} \int_1^3 f(x) dx = \frac{1}{2} \int_1^3 f(x) dx$$<br />We use integration by parts to evaluate $\int_1^3 xf'(x) dx$. Let $u=x$ and $dv = f'(x)dx$. Then $du = dx$ and $v = f(x)$.<br />Using integration by parts, we have:<br />$$\int_1^3 xf'(x) dx = xf(x) \Big|_1^3 - \int_1^3 f(x) dx = 3f(3) - 1f(1) - \int_1^3 f(x) dx$$<br />We are given that $\int_1^3 xf'(x) dx = 13$, $f(1) = 3$, and $f(3) = 1$. Substituting these values, we get:<br />$$13 = 3(1) - 1(3) - \int_1^3 f(x) dx$$<br />$$13 = 0 - \int_1^3 f(x) dx$$<br />$$\int_1^3 f(x) dx = -13$$<br />Therefore, the average value of $f(x)$ on $[1,3]$ is:<br />$$A = \frac{1}{2} \int_1^3 f(x) dx = \frac{1}{2}(-13) = -\frac{13}{2}$$<br /><br />Thus, the average value of $f$ on the interval $[1,3]$ is $-\frac{13}{2}$.<br />