int_0^1 {(1 - x)^9} dx = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $I = \int_0^1 {(1 - x)^9} dx$. Using the substitution method,let $u = 1 - x$,then $du = -dx$,which implies $dx = -du$. When $x = 0$,$u = 1$. W

Vedclass · Accepted Answer

$\frac{1}{10}$

Vedclass · Answer

(B) Let $I = \int_0^1 {(1 - x)^9} dx$. Using the substitution method,let $u = 1 - x$,then $du = -dx$,which implies $dx = -du$. When $x = 0$,$u = 1$. When $x = 1$,$u = 0$. Substituting these into the integral: $I = \int_1^0 u^9 (-du) = \int_0^1 u^9 du$. Evaluating the integral: $I = \left[ \frac{u^{10}}{10} \right]_0^1 = \frac{1^{10}}{10} - \frac{0^{10}}{10} = \frac{1}{10} - 0 = \frac{1}{10}$.

$\int_0^1 {(1 - x)^9} dx = $

Similar Questions

The area of the region under the graph of $y = xe^{-ax}$ as $x$ varies from $0$ to $\infty$,where $'a'$ is a positive constant,is

$ \int_{-\pi / 4}^{\pi / 4} \frac{dx}{1+\cos 2x} $ is equal to

If $\int_a^b x^3 dx = 0$ and $\int_a^b x^2 dx = \frac{2}{3}$,then

$\int_{a}^{b} \operatorname{sgn}(x) \, dx = \dots$ (where $a, b \in \mathbb{R}$)

The integral $\int_{7\pi/4}^{7\pi/3} \sqrt{\tan^2 x} \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module