I = int_{0}^{1} x left| x - frac{1}{2} right| | vedclass

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

Question

(C) આપેલ છે $I = \int_{0}^{1} x \left| x - \frac{1}{2} \right| dx$. અહીં માનાંક વિધેય હોવાથી,આપણે સંકલનને $x = 1/2$ બિંદુએ વિભાજિત કરીશું: $I = \int_{

Vedclass · Accepted Answer

$1/8$

Vedclass · Answer

(C) આપેલ છે $I = \int_{0}^{1} x \left| x - \frac{1}{2} \right| dx$. અહીં માનાંક વિધેય હોવાથી,આપણે સંકલનને $x = 1/2$ બિંદુએ વિભાજિત કરીશું: $I = \int_{0}^{1/2} x \left( -\left( x - \frac{1}{2} \right) \right) dx + \int_{1/2}^{1} x \left( x - \frac{1}{2} \right) dx$. $I = \int_{0}^{1/2} \left( \frac{1}{2}x - x^2 \right) dx + \int_{1/2}^{1} \left( x^2 - \frac{1}{2}x \right) dx$. સંકલન કરતા: $I = \left[ \frac{x^2}{4} - \frac{x^3}{3} \right]_{0}^{1/2} + \left[ \frac{x^3}{3} - \frac{x^2}{4} \right]_{1/2}^{1}$. $I = \left( \frac{(1/2)^2}{4} - \frac{(1/2)^3}{3} \right) + \left( \left( \frac{1}{3} - \frac{1}{4} \right) - \left( \frac{(1/2)^3}{3} - \frac{(1/2)^2}{4} \right) \right)$. $I = \left( \frac{1}{16} - \frac{1}{24} \right) + \left( \frac{1}{12} - \left( \frac{1}{24} - \frac{1}{16} \right) \right)$. $I = \left( \frac{3-2}{48} \right) + \left( \frac{1}{12} - \left( \frac{2-3}{48} \right) \right) = \frac{1}{48} + \frac{1}{12} + \frac{1}{48} = \frac{2}{48} + \frac{4}{48} = \frac{6}{48} = \frac{1}{8}$.

$t \ (\text{સેકન્ડમાં})$	$0$	$2$	$4$	$6$	$8$	$10$
$v \ (\text{m/s માં})$	$0$	$12$	$16$	$20$	$35$	$60$

$I = \int_{0}^{1} x \left| x - \frac{1}{2} \right| dx$ નું મૂલ્ય શોધો.

Similar Questions

સાબિત કરો કે $\int_{0}^{1} \sin^{-1} x \, dx = \frac{\pi}{2} - 1$.

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

જો $I=\int_1^3 \sqrt{3+x+x^2} dx$ હોય,તો $I$ કયા અંતરાલમાં આવે છે?

$\int_{2}^{3} \frac{dx}{x^2 - x} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module