CUET UG 2025 — Mathematics Calculus

Given: $\displaystyle \int_{-\pi/2}^{\pi/2}(x^5 + x^3\cos x)dx$∫−π/2π/2​(x5+x3cosx)dx

For any odd function $f(x)$f(x) where $f(-x) = -f(x)$f(−x)=−f(x):

$$\int_{-a}^{a}f(x)dx = 0$$∫−aa​f(x)dx=0

The negative side cancels the positive side.

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Breaking the integral into two parts:

$$\int_{-\pi/2}^{\pi/2}(x^5 + x^3\cos x)dx = \int_{-\pi/2}^{\pi/2}x^5 dx + \int_{-\pi/2}^{\pi/2}x^3\cos x \, dx$$∫−π/2π/2​(x5+x3cosx)dx=∫−π/2π/2​x5dx+∫−π/2π/2​x3cosxdx

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Consider $f(x) = x^5$f(x)=x5

$$f(-x) = (-x)^5$$f(−x)=(−x)5

$$f(-x) = -x^5$$f(−x)=−x5

$$f(-x) = -f(x)$$f(−x)=−f(x)

Therefore $x^5$x5 is an odd function.

$$\int_{-\pi/2}^{\pi/2}x^5 dx = 0$$∫−π/2π/2​x5dx=0

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Consider $g(x) = x^3\cos x$g(x)=x3cosx

$$g(-x) = (-x)^3 \cdot \cos(-x)$$g(−x)=(−x)3⋅cos(−x)

$$g(-x) = -x^3 \cdot \cos x$$g(−x)=−x3⋅cosx

Since $\cos(-x) = \cos x$cos(−x)=cosx:

$$g(-x) = -g(x)$$g(−x)=−g(x)

Therefore $x^3\cos x$x3cosx is an odd function.

$$\int_{-\pi/2}^{\pi/2}x^3\cos x \, dx = 0$$∫−π/2π/2​x3cosxdx=0

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$$\int_{-\pi/2}^{\pi/2}(x^5 + x^3\cos x)dx = 0 + 0$$∫−π/2π/2​(x5+x3cosx)dx=0+0

$$\int_{-\pi/2}^{\pi/2}(x^5 + x^3\cos x)dx = 0$$∫−π/2π/2​(x5+x3cosx)dx=0