int_2^5 (sqrt{x+2 sqrt{x-1}} + sqrt{x-2 sqrt{ | vedclass

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

Question

(C) ધારો કે $I = \int_2^5 (\sqrt{x+2 \sqrt{x-1}} + \sqrt{x-2 \sqrt{x-1}}) dx$. વર્ગમૂળની અંદરના પદોને પૂર્ણ વર્ગ તરીકે લખતા: $x + 2\sqrt{x-1} = (\sqrt

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(C) ધારો કે $I = \int_2^5 (\sqrt{x+2 \sqrt{x-1}} + \sqrt{x-2 \sqrt{x-1}}) dx$. વર્ગમૂળની અંદરના પદોને પૂર્ણ વર્ગ તરીકે લખતા: $x + 2\sqrt{x-1} = (\sqrt{x-1})^2 + 2\sqrt{x-1} + 1 = (\sqrt{x-1} + 1)^2$. $x - 2\sqrt{x-1} = (\sqrt{x-1})^2 - 2\sqrt{x-1} + 1 = (\sqrt{x-1} - 1)^2$. તેથી,સંકલન આ મુજબ થશે: $I = \int_2^5 (\sqrt{(\sqrt{x-1} + 1)^2} + \sqrt{(\sqrt{x-1} - 1)^2}) dx$. અહીં $x \in [2, 5]$ હોવાથી,$\sqrt{x-1} \ge 1$,તેથી $\sqrt{x-1} - 1 \ge 0$. $I = \int_2^5 (\sqrt{x-1} + 1 + \sqrt{x-1} - 1) dx = \int_2^5 2\sqrt{x-1} dx$. $I = 2 \int_2^5 (x-1)^{1/2} dx = 2 \left[ \frac{(x-1)^{3/2}}{3/2} ight]_2^5 = 2 	imes \frac{2}{3} [(x-1)^{3/2}]_2^5$. $I = \frac{4}{3} [(5-1)^{3/2} - (2-1)^{3/2}] = \frac{4}{3} [4^{3/2} - 1^{3/2}] = \frac{4}{3} [8 - 1] = \frac{4}{3} 	imes 7 = \frac{28}{3}$.

$\int_2^5 (\sqrt{x+2 \sqrt{x-1}} + \sqrt{x-2 \sqrt{x-1}}) dx = $ ($/3$ માં)

Similar Questions

$I = \int_{\pi/4}^{\pi/3} \frac{8 \sin x - \sin 2x}{x} dx$. તો

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

$\int_0^{\pi /8} \cos^3(4\theta) \, d\theta = $

$\int_{-4}^5 \frac{1}{\sqrt{20+x-x^2}} dx=$

$\int_0^1 x \left|x - \frac{1}{2}\right| dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module