CUET UG 2025 — Mathematics Algebra

The region is enclosed by the curve $y = \cos x$y=cosx, the x-axis, and the vertical lines $x = \frac{-\pi}{2}$x=2−π​ and $x = \frac{\pi}{2}$x=2π​.

For this interval:

• At $x = \frac{-\pi}{2}$x=2−π​: $\cos(\frac{-\pi}{2}) = 0$cos(2−π​)=0
• At $x = 0$x=0: $\cos(0) = 1$cos(0)=1
• At $x = \frac{\pi}{2}$x=2π​: $\cos(\frac{\pi}{2}) = 0$cos(2π​)=0

The curve $y = \cos x$y=cosx is always positive (above the x-axis) in the interval $[\frac{-\pi}{2}, \frac{\pi}{2}]$[2−π​,2π​].

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The area under the curve is given by:

Area $\displaystyle = \int_{-\pi/2}^{\pi/2} \cos x \, dx$=∫−π/2π/2​cosxdx

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The antiderivative of $\cos x$cosx is $\sin x$sinx:

$$\int \cos x \, dx = \sin x + C$$∫cosxdx=sinx+C

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Evaluating the definite integral:

Area $= [\sin x]_{-\pi/2}^{\pi/2}$=[sinx]−π/2π/2​

Area $= \sin(\frac{\pi}{2}) - \sin(\frac{-\pi}{2})$=sin(2π​)−sin(2−π​)

Area $= 1 - (-1)$=1−(−1)

Area $= 1 + 1$=1+1

Area $= 2$=2

Therefore, the area of the region is $2$2 square units.