CUET UG 2025 — Mathematics Calculus

When integrating a product of different types of functions like $x$x and $e^x$ex, use integration by parts.

$$\int u \, dv = uv - \int v \, du$$∫udv=uv−∫vdu

Using the ILATE rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential), choose the function that appears first as $u$u.

Let $u = x$u=x (Algebraic) and $dv = e^x dx$dv=exdx (Exponential)

Then $du = dx$du=dx and $v = e^x$v=ex

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Applying integration by parts:

$$\int x e^x dx = x \cdot e^x - \int e^x \cdot dx$$∫xexdx=x⋅ex−∫ex⋅dx

$$= x e^x - e^x + C$$=xex−ex+C

$$= e^x(x - 1) + C$$=ex(x−1)+C

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Evaluating the definite integral from 0 to 1:

$$\int_0^1 x e^x dx = \left[e^x(x - 1)\right]_0^1$$∫01​xexdx=[ex(x−1)]01​

At $x = 1$x=1:
$$e^1(1 - 1) = 0$$e1(1−1)=0

At $x = 0$x=0:
$$e^0(0 - 1) = -1$$e0(0−1)=−1

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$$\int_0^1 x e^x dx = 0 - (-1)$$∫01​xexdx=0−(−1)

$$= 1$$=1

Therefore, the value is $1$1.