Calculus

Integrals

Easy

common

The value of $\displaystyle\int_0^1 xe^{x^2}\,dx$ is equal to :

$e+1$

$\dfrac{e+1}{2}$

$e-1$

$\dfrac{e-1}{2}$

✅ Correct Option: 4

Put $t = x^2$ , so $dt = 2x\,dx$ and the limits become $0$ to $1$ .

The integral becomes $\dfrac{1}{2}\int_0^1 e^t dt = \dfrac{1}{2}\left[e^t\right]_0^1 = \dfrac{e-1}{2}$ .

2025: 27 May Shift 1

The value of $\int_{-1}^{1} |x^3 - x| dx$ is

2025: 3 June Shift 1

The value of $\int \frac{(x^4 - x)^{1/4}}{x^5} dx$ is equal to (where C is an arbitrary constant)

2025: 3 June Shift 2

Match List-I with List-II

List-I	List-II
(A) $\int \dfrac{dx}{x^2 - 16}$	(I) $\dfrac{1}{8} \log \left\vert \dfrac{4 + x}{4 - x} \right\vert + c$ , Where C is an arbitrary constant,
(B) $\int \dfrac{dx}{x^2 + 16}$	(II) $\log \left\vert x + \sqrt{x^2 - 16} \right\vert + c$ , Where C is an arbitrary constant,
(C) $\int \dfrac{dx}{16 - x^2}$	(III) $\dfrac{1}{8} \log \left\vert \dfrac{x - 4}{x + 4} \right\vert + c$ , Where C is an arbitrary constant,
(D) $\int \dfrac{dx}{\sqrt{x^2 - 16}}$	(IV) $\dfrac{1}{4} \tan^{-1} \left( \dfrac{x}{4} \right) + c$ , Where C is an arbitrary constant,

Choose the correct answer from the options given below:

2026: 18 May Shift 2

$\int \dfrac{e^{7\log x}-e^{6\log x}}{e^{5\log x}-e^{4\log x}}\,dx$ , $x>0$ and $x\ne 1$ is equal to (consider $\log x = \log_e x$ )

2025: 26 May Shift 2

$\int \frac{(x-1)e^x}{x^2} dx, x > 0$ equals (where C is an arbitrary constant)

2025: 13 May Shift 2

$\int_{-\frac{5}{2}}^{\frac{5}{2}} |x| dx$ is equal to

The value of ∫01xex2 dx\displaystyle\int_0^1 xe^{x^2}\,dx∫01​xex2dx is equal to :