If int_0^1 frac{1}{sqrt{3+x}+sqrt{1+x}} d x=a | vedclass

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

Question

(D) Rationalize the integrand: $\int_0^1 \frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)} d x = \frac{1}{2} \int_0^1 (\sqrt{3+x}-\sqrt{1+x}) d x$ Evaluate th

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Rationalize the integrand: $\int_0^1 \frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)} d x = \frac{1}{2} \int_0^1 (\sqrt{3+x}-\sqrt{1+x}) d x$ Evaluate the integral: $\frac{1}{2} \left[ \frac{2}{3}(3+x)^{3/2} - \frac{2}{3}(1+x)^{3/2} \right]_0^1$ $= \frac{1}{3} \left[ (3+x)^{3/2} - (1+x)^{3/2} \right]_0^1$ $= \frac{1}{3} \left[ (4^{3/2} - 2^{3/2}) - (3^{3/2} - 1^{3/2}) \right]$ $= \frac{1}{3} \left[ (8 - 2\sqrt{2}) - (3\sqrt{3} - 1) \right] = \frac{1}{3} [9 - 2\sqrt{2} - 3\sqrt{3}] = 3 - \frac{2}{3}\sqrt{2} - \sqrt{3}$ Comparing with $a+b\sqrt{2}+c\sqrt{3}$,we get $a=3$,$b=-\frac{2}{3}$,$c=-1$ Calculate $2a+3b-4c = 2(3) + 3(-\frac{2}{3}) - 4(-1) = 6 - 2 + 4 = 8$

If $\int_0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}} d x=a+b \sqrt{2}+c \sqrt{3}$,where $a, b, c$ are rational numbers,then $2 a+3 b-4 c$ is equal to :

Similar Questions

$4 \int_0^1 \left(\frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}}\right) dx - 3 \log_e(\sqrt{3})$ is equal to :

If $f(x) = \operatorname{Max}\{x^3-4, x^4-4\}$ and $g(x) = \operatorname{Min}\{x^2, x^3\}$,then $\int_{-1}^1 (f(x) - g(x)) \, dx =$

$\int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{1+\sin ^{4} x} d x=$

$\int_{0}^{\pi /2}{\frac{dx}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}}\,=$

$\int_0^\infty {\frac{{x\,dx}}{{(1 + x)(1 + {x^2})}}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module