CUET UG 2025 — Mathematics Calculus

The function is $(x^7 + x^5 + x^3 + x + 1)$(x7+x5+x3+x+1) with limits from $-1$−1 to $1$1 (symmetric around zero).

The function can be split into:

• Odd power terms: $x^7, x^5, x^3, x$x7,x5,x3,x
• Even power term: $1$1

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When integrating an odd function over symmetric limits (like $-1$−1 to $1$1), the result is always zero because the negative side perfectly cancels the positive side.

$$\int_{-1}^{1}x^7 dx = 0$$∫−11​x7dx=0

$$\int_{-1}^{1}x^5 dx = 0$$∫−11​x5dx=0

$$\int_{-1}^{1}x^3 dx = 0$$∫−11​x3dx=0

$$\int_{-1}^{1}x dx = 0$$∫−11​xdx=0

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The constant term remains:

$$\int_{-1}^{1}1 \, dx = [x]_{-1}^{1}$$∫−11​1dx=[x]−11​

$$= (1) - (-1)$$=(1)−(−1)

$$= 2$$=2

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$$\int_{-1}^{1}(x^7 + x^5 + x^3 + x + 1)dx = 0 + 0 + 0 + 0 + 2$$∫−11​(x7+x5+x3+x+1)dx=0+0+0+0+2

$$= 2$$=2