int(3-x) sqrt{4-x} , dx = (Where C is a cons | vedclass

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

Question

(C) Let $I = \int(3-x) \sqrt{4-x} \, dx$. Substitute $u = 4-x$,then $du = -dx$,which means $dx = -du$. Also,$x = 4-u$,so $3-x = 3-(4-u) = u-1$. Substi

Vedclass · Accepted Answer

$\frac{2}{3}(4-x)^{3/2} - \frac{2}{5}(4-x)^{5/2} + C$

Vedclass · Answer

(C) Let $I = \int(3-x) \sqrt{4-x} \, dx$. Substitute $u = 4-x$,then $du = -dx$,which means $dx = -du$. Also,$x = 4-u$,so $3-x = 3-(4-u) = u-1$. Substituting these into the integral: $I = \int (u-1) \sqrt{u} (-du) = \int (1-u) \sqrt{u} \, du$ $I = \int (u^{1/2} - u^{3/2}) \, du$ $I = \frac{u^{3/2}}{3/2} - \frac{u^{5/2}}{5/2} + C$ $I = \frac{2}{3} u^{3/2} - \frac{2}{5} u^{5/2} + C$ Substituting $u = 4-x$ back: $I = \frac{2}{3} (4-x)^{3/2} - \frac{2}{5} (4-x)^{5/2} + C$.

$\int(3-x) \sqrt{4-x} \, dx = $ (Where $C$ is a constant of integration.)

Similar Questions

The integral $\int \frac{(x^8-x^2) dx}{(x^{12}+3x^6+1) \tan^{-1}(x^3+\frac{1}{x^3})}$ is equal to:

$\int {{e^{3\log x}}{{({x^4} + 1)}^{ - 1}}\,dx} = $

The value of $\int \frac{e^x}{e^x + 1} \,dx$ is

$\int (\sqrt{\tan x} + \sqrt{\cot x}) \, dx$ is equal to-

The integral $\int \frac{1}{\sqrt[4]{(x-1)^{3}(x+2)^{5}}} dx$ is equal to : (where $C$ is a constant of integration)

Vedclass Test Series

Exam Paper Generator

Online Exam Module