The value of int_{2}^{8} frac{sqrt{10-x}}{sq | vedclass

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

Question

(D) Let $ I = \int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x \quad (1) $ Using the property $ \int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x)

Vedclass · Accepted Answer

$ 03 $

Vedclass · Answer

(D) Let $ I = \int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x \quad (1) $ Using the property $ \int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x $,where $ a=2 $ and $ b=8 $,we have $ a+b-x = 2+8-x = 10-x $. Substituting this into the integral,we get: $ I = \int_{2}^{8} \frac{\sqrt{10-(10-x)}}{\sqrt{10-x}+\sqrt{10-(10-x)}} d x $ $ I = \int_{2}^{8} \frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}} d x \quad (2) $ Adding equations $(1)$ and $(2)$: $ 2I = \int_{2}^{8} \frac{\sqrt{10-x}+\sqrt{x}}{\sqrt{x}+\sqrt{10-x}} d x $ $ 2I = \int_{2}^{8} 1 d x $ $ 2I = [x]_{2}^{8} = 8 - 2 = 6 $ $ I = \frac{6}{2} = 3 $

The value of $ \int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x $ is

Similar Questions

$\int_5^9 \frac{\log_3 x^2}{\log_3 x^2 + \log_3(588 - 84x + 3x^2)} dx =$

$\int_{\frac{1}{2}}^2 \frac{1}{x} \operatorname{cosec}^{101}\left(x-\frac{1}{x}\right) d x=$

$\int_{0}^{\pi} x f(\sin x) dx = $

$\int_0^{2 \pi} \sin ^3 x \cos ^2 x \, dx = $ . . . . . . .

The value of $\int_1^3 \sqrt{3 + x^3} \,dx$ lies in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module