JEE Advanced 2024 Paper 2, Mathematics Section 4 Q4: Area as Definite Integral

JEE Advanced2024Paper 2Mathematics Section 4

PARAGRAPH II

Let $f : [0, \frac{π}{2}] \to [0, 1]$ be the function defined by $f (x) = sin^{2} x$ and let $g : [0, \frac{π}{2}] \to [0, \infty)$ be the function defined by $g (x) = \frac{π x}{2} - x^{2}$ .

(There are two questions based on PARAGRAPH “II”, the question given below is one of them)

Q2. The value of $\frac{16}{π ^{3}} \int_{0}^{π /2} f (x) g (x) d x$ is ______.

Your answer

Solution

From the previous question, $2 \int_{0}^{π /2} f (x) g (x) d x = \int_{0}^{π /2} g (x) d x$ . So it suffices to evaluate the integral of $g$ .

$g (x) = \frac{π x}{2} - x^{2} = (\frac{π}{4})^{2} - (x - \frac{π}{4})^{2} .$

This describes a semicircle of radius $\frac{π}{4}$ centred at $x = \frac{π}{4}$ on the x-axis. The integral $\int_{0}^{π /2} g (x) d x$ equals the area of a semicircular disc of radius $\frac{π}{4}$ :

$\int_{0}^{π /2} g (x) d x = \frac{1}{2} π (\frac{π}{4})^{2} = \frac{π ^{3}}{32} .$

Therefore $\int_{0}^{π /2} f (x) g (x) d x = \frac{1}{2} \cdot \frac{π ^{3}}{32} = \frac{π ^{3}}{64} .$

Multiplying by $\frac{16}{π ^{3}}$ :

$\frac{16}{π ^{3}} \cdot \frac{π ^{3}}{64} = \frac{1}{4} = 0.25.$

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JEE Advanced 2024 Paper 2, Mathematics Section 4 Q4: Area as Definite Integral

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