Evaluate the definite integral: int_0^1 frac{ | vedclass

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

Question

(B) Let $I = \int_0^1 \frac{dx}{\sqrt{1+x} - \sqrt{x}}$.Rationalizing the denominator,we multiply the numerator and denominator by $(\sqrt{1+x} + \sqr

Vedclass · Accepted Answer

$\frac{4\sqrt{2}}{3}$

Vedclass · Answer

(B) Let $I = \int_0^1 \frac{dx}{\sqrt{1+x} - \sqrt{x}}$.Rationalizing the denominator,we multiply the numerator and denominator by $(\sqrt{1+x} + \sqrt{x})$:$I = \int_0^1 \frac{\sqrt{1+x} + \sqrt{x}}{(1+x) - x} dx = \int_0^1 (\sqrt{1+x} + \sqrt{x}) dx$.Integrating term by term:$I = \left[ \frac{2}{3}(1+x)^{3/2} + \frac{2}{3}x^{3/2} \right]_0^1$.Evaluating at the limits:$I = \left( \frac{2}{3}(2)^{3/2} + \frac{2}{3}(1)^{3/2} \right) - \left( \frac{2}{3}(1)^{3/2} + 0 \right)$.$I = \frac{2}{3}(2\sqrt{2}) + \frac{2}{3} - \frac{2}{3} = \frac{4\sqrt{2}}{3}$.

Evaluate the definite integral: $\int_0^1 \frac{dx}{\sqrt{1+x} - \sqrt{x}}$

Similar Questions

$\int_0^{\pi / 2} \sin ^5\left(\frac{x}{2}\right) \cdot \sin x \, dx =$

The value of the integral $\int_{\pi/6}^{\pi/3} \left(\frac{4 - \csc^2 x}{\cos^4 x}\right) dx$ is:

For any real number $x$,let $[x]$ denote the largest integer less than or equal to $x$. If $I = \int_0^{10} \left[ \sqrt{\frac{10x}{x+1}} \right] dx$,then the value of $9I$ is . . . . . .

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot ^9 x \, dx =$

Evaluate the definite integral: $\int_{0}^{a} \sqrt{\frac{x}{a-x}} \, dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module