CUET UG 2022 — Mathematics Applied-Mathematics

We need the area under $y = x^2$y=x2, above the $x$x-axis, from $x = 0$x=0 to $x = \pi$x=π.

Whenever we need the area under a curve, we use definite integration — think of it as summing up infinitely thin vertical strips between the curve and the $x$x-axis.

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Setting up the integral:

$$\text{Area} = \int_0^{\pi} x^2 \, dx$$Area=∫0π​x2dx

Since both $x$x and $y = x^2$y=x2 are non-negative between $0$0 and $\pi$π, we are entirely in the first quadrant.

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Applying the power rule $\displaystyle \int x^n \, dx = \dfrac{x^{n+1}}{n+1} + C$∫xndx=n+1xn+1​+C :

$$\int_0^{\pi} x^2 \, dx = \left[\dfrac{x^3}{3}\right]_0^{\pi}$$∫0π​x2dx=[3x3​]0π​

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Substituting the upper and lower limits:

$$= \dfrac{(\pi)^3}{3} - \dfrac{(0)^3}{3}$$=3(π)3​−3(0)3​

$$= \dfrac{\pi^3}{3} - 0$$=3π3​−0

$$= \boxed{\dfrac{\pi^3}{3}}$$=3π3​​

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Don't confuse this with area under $y = \sin x$y=sinx or $y = \cos x$y=cosx which also commonly use $0$0 to $\pi$π as limits. Here the curve is $y = x^2$y=x2, a basic polynomial, so we simply apply the power rule — no trigonometry involved.