CUET UG 2025 — Mathematics Calculus

$$\int_{1}^{e} \frac{\log x}{x}\, dx$$∫1e​xlogx​dx

Let $t = \log x \implies dt = \frac{1}{x}\,dx$t=logx⟹dt=x1​dx

When $x = 1$x=1, $t = 0$t=0 and when $x = e$x=e, $t = 1$t=1

$$\int_{0}^{1} t\, dt = \left[\frac{t^2}{2}\right]_0^1$$∫01​tdt=[2t2​]01​

$$= \frac{1}{2} - 0$$=21​−0

$$= \frac{1}{2}$$=21​

(A) → (II)

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$$\int_{-2}^{2} x^3(1 - x^2)\, dx$$∫−22​x3(1−x2)dx

Let $f(x) = x^3(1 - x^2) = x^3 - x^5$f(x)=x3(1−x2)=x3−x5

$$f(-x) = (-x)^3 - (-x)^5 = -x^3 + x^5 = -f(x)$$f(−x)=(−x)3−(−x)5=−x3+x5=−f(x)

Since $f(x)$f(x) is an odd function, $\displaystyle \int_{-a}^{a} f(x)\,dx = 0$∫−aa​f(x)dx=0

$$\int_{-2}^{2} x^3(1 - x^2)\, dx = 0$$∫−22​x3(1−x2)dx=0

(B) → (III)

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$$\int_{1}^{2} x\, dx$$∫12​xdx

$$= \left[\frac{x^2}{2}\right]_1^2$$=[2x2​]12​

$$= \frac{4}{2} - \frac{1}{2}$$=24​−21​

$$= \frac{3}{2}$$=23​

(C) → (IV)

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$$\int_{-2}^{2} |x|\, dx$$∫−22​∣x∣dx

Since $|x|$∣x∣ is an even function, $\displaystyle \int_{-a}^{a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx$∫−aa​f(x)dx=2∫0a​f(x)dx

$$= 2\int_{0}^{2} x\, dx$$=2∫02​xdx

$$= 2\left[\frac{x^2}{2}\right]_0^2$$=2[2x2​]02​

$$= 2\left(\frac{4}{2} - 0\right)$$=2(24​−0)

$$= 4$$=4

(D) → (I)

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List-I	List-II
(A)	(II) $\frac{1}{2}$21​
(B)	(III) $0$0
(C)	(IV) $\frac{3}{2}$23​
(D)	(I) $4$4