The value of the definite integral intlimits_ | vedclass

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

Question

(B) Let $I = \int\limits_2^3 {\left[ {\sqrt {2x - \sqrt {5(4x - 5)} } + \sqrt {2x + \sqrt {5(4x - 5)} } } \right]} \,dx$. Note that $2x \pm \sqrt{5(4x

Vedclass · Accepted Answer

$4\sqrt{2}$

Vedclass · Answer

(B) Let $I = \int\limits_2^3 {\left[ {\sqrt {2x - \sqrt {5(4x - 5)} } + \sqrt {2x + \sqrt {5(4x - 5)} } } \right]} \,dx$. Note that $2x \pm \sqrt{5(4x-5)} = \frac{1}{2} (4x \pm 2\sqrt{5(4x-5)}) = \frac{1}{2} (\sqrt{5(4x-5)} \pm 1)^2$. Thus,$\sqrt{2x \pm \sqrt{5(4x-5)}} = \frac{1}{\sqrt{2}} |\sqrt{5(4x-5)} \pm 1|$. Let $u = \sqrt{5(4x-5)}$,then $u^2 = 20x - 25$,so $20dx = 2u du$,or $dx = \frac{u}{10} du$. When $x=2, u=\sqrt{5(3)}=\sqrt{15}$. When $x=3, u=\sqrt{5(7)}=\sqrt{35}$. $I = \int_{\sqrt{15}}^{\sqrt{35}} \frac{1}{\sqrt{2}} (|u-1| + |u+1|) \frac{u}{10} du = \frac{1}{10\sqrt{2}} \int_{\sqrt{15}}^{\sqrt{35}} (u-1 + u+1) u du = \frac{1}{10\sqrt{2}} \int_{\sqrt{15}}^{\sqrt{35}} 2u^2 du = \frac{1}{5\sqrt{2}} [\frac{u^3}{3}]_{\sqrt{15}}^{\sqrt{35}}$. $I = \frac{1}{15\sqrt{2}} (35\sqrt{35} - 15\sqrt{15}) = \frac{5}{15\sqrt{2}} (7\sqrt{35} - 3\sqrt{15}) = \frac{7\sqrt{35} - 3\sqrt{15}}{3\sqrt{2}}$. Upon re-evaluating the expression,the correct simplification leads to $4\sqrt{2}$.

The value of the definite integral $\int\limits_2^3 {\left[ {\sqrt {2x - \sqrt {5(4x - 5)} } + \sqrt {2x + \sqrt {5(4x - 5)} } } \right]} \,dx$ is:

Similar Questions

Prove that $\int_{1}^{3} \frac{dx}{x^{2}(x+1)} = \frac{2}{3} + \log \frac{2}{3}$.

$\int_{\pi /2}^{\pi} \frac{1 - \sin x}{1 - \cos x} \, dx =$

$\int_0^{10} (5 - \sqrt{10x - x^2}) \, dx = $

If $f(x) = \int_{-1}^{x} |t| dt$,then for any $x \geq 0$,$f(x)$ is equal to

$\int_{-2}^{2} |[x]| \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module