Question ID: #487
Let $f:R\rightarrow R$ be a twice differentiable function such that $f(2)=1$. If $F(x)=xf(x)$ for all $x\in R$, $\int_{0}^{2}xF'(x)dx=6$ and $\int_{0}^{2}x^{2}F”(x)dx=40$, then $F'(2)+\int_{0}^{2}F(x)dx$ is equal to:
- (1) 11
- (2) 15
- (3) 9
- (4) 13
Solution:
Given $F(x) = xf(x)$. Since $f(2)=1$, $F(2) = 2f(2) = 2(1) = 2$.
Consider the second integral:
$$\int_{0}^{2} x^{2}F”(x) dx = 40$$
Apply integration by parts ($u=x^2, dv=F”(x)dx$):
$$[x^2 F'(x)]_{0}^{2} – \int_{0}^{2} 2x F'(x) dx = 40$$
$$4F'(2) – 0 – 2\int_{0}^{2} xF'(x) dx = 40$$
We are given $\int_{0}^{2} xF'(x) dx = 6$. Substitute this value:
$$4F'(2) – 2(6) = 40$$
$$4F'(2) – 12 = 40$$
$$4F'(2) = 52 \Rightarrow F'(2) = 13$$
Now consider the first integral $\int_{0}^{2} xF'(x) dx = 6$.
Apply integration by parts ($u=x, dv=F'(x)dx$):
$$[xF(x)]_{0}^{2} – \int_{0}^{2} 1 \cdot F(x) dx = 6$$
$$2F(2) – 0 – \int_{0}^{2} F(x) dx = 6$$
Substitute $F(2) = 2$:
$$2(2) – \int_{0}^{2} F(x) dx = 6$$
$$4 – \int_{0}^{2} F(x) dx = 6$$
$$\int_{0}^{2} F(x) dx = 4 – 6 = -2$$
We need to find $F'(2) + \int_{0}^{2} F(x) dx$:
$$= 13 + (-2)$$
$$= 11$$
Ans. (1)
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