JEE Main 2025 Jan 28 Shift 2, Mathematics Q12: Integration by Parts

JEE Main2025Jan 28, Shift 2Mathematics

Let $f : R \to R$ be a twice differentiable function such that $f (2) = 1$ . If $F (x) = x f (x)$ for all $x \in R$ , $\int_{0}^{2} x F^{'} (x) d x = 6$ and $\int_{0}^{2} x^{2} F^{''} (x) d x = 40$ , then $F^{'} (2) + \int_{0}^{2} F (x) d x$ is equal to :

A
$11$
B
$15$
C
$9$
D
$13$

Solution

Note $F (2) = 2 f (2) = 2$ . Apply integration by parts to $\int_{0}^{2} x F^{'} (x) d x$ :

$[x F (x)]_{0}^{2} - \int_{0}^{2} F (x) d x = 6 \Rightarrow 2 F (2) - \int_{0}^{2} F (x) d x = 6 \Rightarrow \int_{0}^{2} F (x) d x = 4 - 6 = - 2$ .

Now apply integration by parts to $\int_{0}^{2} x^{2} F^{''} (x) d x$ :

$[x^{2} F^{'} (x)]_{0}^{2} - 2 \int_{0}^{2} x F^{'} (x) d x = 40 \Rightarrow 4 F^{'} (2) - 12 = 40 \Rightarrow F^{'} (2) = 13$ .

$∴ F^{'} (2) + \int_{0}^{2} F (x) d x = 13 + (- 2) = 11$

Concept behind this question

Integration by PartsNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Jan 28 Shift 2, Mathematics Q12: Integration by Parts

Concept behind this question

More previous year questions on Integration by Parts

Practice more Mathematics