Question:

Step 1: Find $f'(x)$. Given $f(x) = \int_x^2 \frac{1}{1+y^3} dy$. Applying the Leibniz integral rule, $f'(x) = -\frac{1}{1+x^3}$. Step 2: Apply integration by parts to $\int_0^2 xf(x) dx$. Let $u = f(x)$ and $dv = x dx$. Then $du = f'(x) dx$ and $v = \frac{x^2}{2}$. The integration by parts formula is $\int u dv = uv - \int v du$. $$ \int_0^2 xf(x) dx = \left[\frac{x^2}{2}f(x)\right]_0^2 - \int_0^2 \frac{x^2}{2}f'(x) dx $$ Step 3: Evaluate the boundary term. First, evaluate $f(2)$ and $f(0)$: $$ f(2) = \int_2^2 \frac{1}{1+y^3} dy = 0 $$ $$ f(0) = \int_0^2 \frac{1}{1+y^3} dy $$ Now, substitute these into the boundary term: $$ \left[\frac{x^2}{2}f(x)\right]_0^2 = \frac{2^2}{2}f(2) - \frac{0^2}{2}f(0) = 2 \cdot 0 - 0 \cdot f(0) = 0 $$ Step 4: Evaluate the remaining integral. Substitute $f'(x) = -\frac{1}{1+x^3}$ into the integral: $$ -\int_0^2 \frac{x^2}{2}f'(x) dx = -\int_0^2 \frac{x^2}{2}\left(-\frac{1}{1+x^3}\right) dx = \frac{1}{2}\int_0^2 \frac{x^2}{1+x^3} dx $$ Step 5: Evaluate the integral using substitution. Let $u = 1+x^3$. Then $du = 3x^2 dx$, which implies $x^2 dx = \frac{1}{3} du$. Change the limits of integration: When $x = 0$, $u = 1+0^3 = 1$. When $x = 2$, $u = 1+2^3 = 9$. Substitute these into the integral: $$ \frac{1}{2}\int_1^9 \frac{1}{u} \frac{1}{3} du = \frac{1}{6}\int_1^9 \frac{1}{u} du $$ $$ = \frac{1}{6}[\ln|u|]_1^9 = \frac{1}{6}(\ln 9 - \ln 1) = \frac{1}{6}\ln 9 $$ Step 6: Combine the results. $$ \int_0^2 xf(x) dx = 0 + \frac{1}{6}\ln 9 = \frac{1}{6}\ln 9 $$ This can be written as: $$ \frac{1}{6}\ln 9 = \frac{1}{6}\ln(3^2) = \frac{2}{6}\ln 3 = \frac{1}{3}\ln 3 $$