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(B) Given $\int_0^2 x F^{\prime}(x) dx = 6$. Using integration by parts: $[x F(x)]_0^2 - \int_0^2 F(x) dx = 6$ $2 F(2) - 0 - \int_0^2 F(x) dx = 6$. Si

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$11$

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(B) Given $\int_0^2 x F^{\prime}(x) dx = 6$. Using integration by parts: $[x F(x)]_0^2 - \int_0^2 F(x) dx = 6$ $2 F(2) - 0 - \int_0^2 F(x) dx = 6$. Since $F(x) = x f(x)$,$F(2) = 2 f(2) = 2(1) = 2$. So,$2(2) - \int_0^2 F(x) dx = 6 \implies \int_0^2 F(x) dx = 4 - 6 = -2$. Now,consider $\int_0^2 x^2 F^{\prime \prime}(x) dx = 40$. Using integration by parts: $[x^2 F^{\prime}(x)]_0^2 - \int_0^2 2x F^{\prime}(x) dx = 40$ $4 F^{\prime}(2) - 2 \int_0^2 x F^{\prime}(x) dx = 40$. Substituting the given value $\int_0^2 x F^{\prime}(x) dx = 6$: $4 F^{\prime}(2) - 2(6) = 40$ $4 F^{\prime}(2) - 12 = 40 4 F^{\prime}(2) = 52 F^{\prime}(2) = 13$. Finally,$F^{\prime}(2) + \int_0^2 F(x) dx = 13 + (-2) = 11$.

Let $f : R \rightarrow R$ be a twice differentiable function such that $f(2)=1$. If $F(x) = x f(x)$ for all $x \in R$,$\int_0^2 x F^{\prime}(x) dx = 6$ and $\int_0^2 x^2 F^{\prime \prime}(x) dx = 40$,then $F^{\prime}(2) + \int_0^2 F(x) dx$ is equal to:

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