Definite Integration
Definite Integration
nta_pyq_2025_apr
Grade 12
Question:
Let $f:\mathbb{R}\to\mathbb{R}$ be a twice differentiable function such that $f(2) = 1$. If $F(x) = xf(x)$ for all $x\in\mathbb{R}$, $\displaystyle\int_0^2 xF'(x)\,dx = 6$ and $\displaystyle\int_0^2 x^2 F''(x)\,dx = 40$, then $F'(2)+\displaystyle\int_0^2 F(x)\,dx$ is equal to:
Step-by-Step Solution
Key Concept: Apply IBP to $\int_0^2 xF'(x)dx$ to express it in terms of $F(2)$ and $\int_0^2 F(x)dx$; then apply IBP again to $\int_0^2 x^2 F''(x)dx$ to find $F'(2)$.
$\int_0^2 xF'(x)dx = \left[xF(x)\right]_0^2 - \int_0^2 F(x)dx = 2F(2) - \int_0^2 F(x)dx = 6$.
Since $F(2) = 2f(2) = 2$: $4 - \int_0^2 F(x)dx = 6 \Rightarrow \int_0^2 F(x)dx = -2$.
$\int_0^2 x^2 F''(x)dx = \left[x^2 F'(x)\right]_0^2 - 2\int_0^2 xF'(x)dx = 4F'(2) - 12 = 40$.
$\Rightarrow F'(2) = 13$.
$$F'(2)+\int_0^2 F(x)dx = 13 + (-2) = 11.$$
Correct Answer: 1