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) Let $I = \int_2^5 (\sqrt{x+2 \sqrt{x-1}} + \sqrt{x-2 \sqrt{x-1}}) dx$. We can rewrite the terms inside the square roots as perfect squares: $x + 2

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(C) Let $I = \int_2^5 (\sqrt{x+2 \sqrt{x-1}} + \sqrt{x-2 \sqrt{x-1}}) dx$. We can rewrite the terms inside the square roots as perfect squares: $x + 2\sqrt{x-1} = (\sqrt{x-1})^2 + 2\sqrt{x-1} + 1 = (\sqrt{x-1} + 1)^2$. $x - 2\sqrt{x-1} = (\sqrt{x-1})^2 - 2\sqrt{x-1} + 1 = (\sqrt{x-1} - 1)^2$. Thus,the integral becomes: $I = \int_2^5 (\sqrt{(\sqrt{x-1} + 1)^2} + \sqrt{(\sqrt{x-1} - 1)^2}) dx$. Since $x \in [2, 5]$,$\sqrt{x-1} \ge 1$,so $\sqrt{x-1} - 1 \ge 0$. $I = \int_2^5 (\sqrt{x-1} + 1 + \sqrt{x-1} - 1) dx = \int_2^5 2\sqrt{x-1} dx$. $I = 2 \int_2^5 (x-1)^{1/2} dx = 2 \left[ \frac{(x-1)^{3/2}}{3/2} ight]_2^5 = 2 	imes \frac{2}{3} [(x-1)^{3/2}]_2^5$. $I = \frac{4}{3} [(5-1)^{3/2} - (2-1)^{3/2}] = \frac{4}{3} [4^{3/2} - 1^{3/2}] = \frac{4}{3} [8 - 1] = \frac{4}{3} 	imes 7 = \frac{28}{3}$.

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

Similar Questions

The value of the integral $\int_{-1}^2 \log _e\left(x+\sqrt{x^2+1}\right) d x$ is:

If $I=\int_{1}^{2} \frac{dx}{\sqrt{2x^{3}-9x^{2}+12x+4}},$ then

If $\frac{d}{dx}\{f(x)\} = g(x)$, then $\int_a^b f(x) g(x) dx$ is equal to

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{{n^2}}}{{\sec }^2}\frac{1}{{{n^2}}} + \frac{2}{{{n^2}}}{{\sec }^2}\frac{4}{{{n^2}}} + ..... + \frac{1}{n}{{\sec }^2}1} \right]$ equals

If $\int_0^b \frac{dx}{1+x^2} = \int_b^{\infty} \frac{dx}{1+x^2}$,then $b$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module