int_{3}^{6} frac{sqrt{x}}{sqrt{9-x} + sqrt{x} | vedclass

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

Question

(B) Let $I = \int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} \, dx \quad \dots(i)$Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(B) Let $I = \int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} \, dx \quad \dots(i)$Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,we get:$I = \int_{3}^{6} \frac{\sqrt{9-x}}{\sqrt{9-(9-x)} + \sqrt{9-x}} \, dx$$I = \int_{3}^{6} \frac{\sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} \, dx \quad \dots(ii)$Adding equations $(i)$ and $(ii)$:$2I = \int_{3}^{6} \frac{\sqrt{x} + \sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} \, dx$$2I = \int_{3}^{6} 1 \, dx$$2I = [x]_{3}^{6} = 6 - 3 = 3$$I = \frac{3}{2}$

$\int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} \, dx = $

Similar Questions

The energy of a photon is calculated by

The set of solutions of the equation $(\sqrt{3}-1) \sin \theta+(\sqrt{3}+1) \cos \theta=2$ is

To which series does $Catharanthus$ $roseus$ belong?

The function $f(x)=\log (1+x)-\frac{2 x}{2+x}$ is increasing on

Vedclass Test Series

Exam Paper Generator

Online Exam Module