CUET UG 2025 — Mathematics Calculus

Let $u = 2-x$u=2−x

Then $x = 2-u$x=2−u

$du = -dx$du=−dx, so $dx = -du$dx=−du

When $x = 0$x=0, $u = 2$u=2

When $x = 2$x=2, $u = 0$u=0

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$$\int_0^2 x(2-x)^n dx = \int_2^0 (2-u) \cdot u^n \cdot (-du)$$∫02​x(2−x)ndx=∫20​(2−u)⋅un⋅(−du)

$$= \int_0^2 (2-u) u^n \, du$$=∫02​(2−u)undu

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$$\int_0^2 (2-u) u^n \, du = \int_0^2 (2u^n - u^{n+1}) \, du$$∫02​(2−u)undu=∫02​(2un−un+1)du

$$= \int_0^2 2u^n \, du - \int_0^2 u^{n+1} \, du$$=∫02​2undu−∫02​un+1du

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$$= 2 \cdot \left[\frac{u^{n+1}}{n+1}\right]_0^2 - \left[\frac{u^{n+2}}{n+2}\right]_0^2$$=2⋅[n+1un+1​]02​−[n+2un+2​]02​

$$= 2 \cdot \frac{2^{n+1}}{n+1} - \frac{2^{n+2}}{n+2}$$=2⋅n+12n+1​−n+22n+2​

$$= \frac{2^{n+2}}{n+1} - \frac{2^{n+2}}{n+2}$$=n+12n+2​−n+22n+2​

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$$= 2^{n+2} \left(\frac{1}{n+1} - \frac{1}{n+2}\right)$$=2n+2(n+11​−n+21​)

$$= 2^{n+2} \left(\frac{(n+2) - (n+1)}{(n+1)(n+2)}\right)$$=2n+2((n+1)(n+2)(n+2)−(n+1)​)

$$= 2^{n+2} \left(\frac{1}{(n+1)(n+2)}\right)$$=2n+2((n+1)(n+2)1​)

$$= \frac{2^{n+2}}{(n+1)(n+2)}$$=(n+1)(n+2)2n+2​