The integral _0^1 x(1-x)^n dx is equal to : Option 1: 1/((n+2)(n+3)) Option 2: 1/((n+1)(n+2)) Option 3: 1/(n(n+1)) Option 4: 1/((n-1)(n-2))

Answer: 1/((n+1)(n+2)). Solution: Using Beta function: _0^1 x^(m-1)(1-x)^(n-1)dx = B(m,n) = ((m-1)!(n-1)!)/((m+n-1)!). Here m=2, integrand exponent = n, so this is B(2, n+1) = (1! n!)/((n+2)!) = 1/((n+1)(n+2)).

Calculus

Integrals

Medium

core

The integral $\int_{0}^{1} x(1-x)^n dx$ is equal to :

$\frac{1}{(n+2)(n+3)}$

$\frac{1}{(n+1)(n+2)}$

$\frac{1}{n(n+1)}$

$\frac{1}{(n-1)(n-2)}$

✅ Correct Option: 2

Using Beta function: $\int_0^1 x^{m-1}(1-x)^{n-1}dx = B(m,n) = \frac{(m-1)!(n-1)!}{(m+n-1)!}$ .

Here $m=2$ , integrand exponent $= n$ , so this is $B(2, n+1) = \frac{1!\,n!}{(n+2)!} = \frac{1}{(n+1)(n+2)}$ .

2025: 15 May Shift 2

$\int \frac{f'(x)}{f(x) \log_e[f(x)]} dx$ is equal to

2026: 21 May Shift 2

$\displaystyle\int e^x(\sec x + \log_e(\sec x + \tan x))dx$ is equal to: (where C is an arbitrary constant)

2026: 22 May Shift 2

Match List-I with List-II

List-I	List-II
(A) $\int_{-1}^{1} \lvert x \rvert dx =$	(I) 3
(B) $\int_{-\pi}^{\pi} \cos x \, dx =$	(II) 1
(C) $\int_{-1}^{1} (\lvert x \rvert-1) dx =$	(III) 0
(D) $\int_{-1}^{1} (\lvert x \rvert+1) dx =$	(IV) -1

Choose the correct answer from the options given below:

2026: 15 May Shift 1

If $\displaystyle\int_0^3 (8x^3+5x^2-2x-k)\,dx = 3$ , where $k$ is constant, then $k$ is equal to

2025: 15 May Shift 2

If $I = \int \frac{x}{x - \sqrt{x^2 - 4}} dx = \alpha x^3 + \beta(x^2 - 4)^{\frac{3}{2}} + \gamma$ , where $\gamma$ is constant of integration, then

2026: 22 May Shift 1

Value of $\int_{-2}^{2}|2x+1|dx$ is:

The integral ∫01x(1−x)ndx\int_{0}^{1} x(1-x)^n dx∫01​x(1−x)ndx is equal to :

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

The integral $\int_{0}^{1} x(1-x)^n dx$ is equal to :