JEE Main 2026 Jan 28 Shift 1, Mathematics Q12: Definite Integration And Fundamental Theorem

JEE Main2026Jan 28, Shift 1Mathematics

Let $f$ be a polynomial function such that $f (x^{2} + 1) = x^{4} + 5 x^{2} + 2$ , for all $x \in R$ . Then $\int_{0}^{3} f (x) d x$ is equal to

A
$\frac{41}{3}$
B
$\frac{33}{2}$
C
$\frac{27}{2}$
D
$\frac{5}{3}$

Solution

Substitute $t = x^{2} + 1$ , so $x^{2} = t - 1$ and $x^{4} = (t - 1)^{2}$ :

$f (t) = (t - 1)^{2} + 5 (t - 1) + 2 = t^{2} - 2 t + 1 + 5 t - 5 + 2 = t^{2} + 3 t - 2$ .

Now integrate:

$\int_{0}^{3} (t^{2} + 3 t - 2) d t = [\frac{t ^{3}}{3} + \frac{3 t ^{2}}{2} - 2 t]_{0}^{3} = 9 + \frac{27}{2} - 6 = 3 + \frac{27}{2} = \frac{33}{2}$ .

Option (2).

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JEE Main 2026 Jan 28 Shift 1, Mathematics Q12: Definite Integration And Fundamental Theorem

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