The value of I = int_{0}^{1} x left| x - frac | vedclass

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

Question

(C) Given $I = \int_{0}^{1} x \left| x - \frac{1}{2} \right| dx$. Since the integrand involves an absolute value,we split the integral at $x = 1/2$: $

Vedclass · Accepted Answer

$1/8$

Vedclass · Answer

(C) Given $I = \int_{0}^{1} x \left| x - \frac{1}{2} \right| dx$. Since the integrand involves an absolute value,we split the integral at $x = 1/2$: $I = \int_{0}^{1/2} x \left( -\left( x - \frac{1}{2} \right) \right) dx + \int_{1/2}^{1} x \left( x - \frac{1}{2} \right) dx$. $I = \int_{0}^{1/2} \left( \frac{1}{2}x - x^2 \right) dx + \int_{1/2}^{1} \left( x^2 - \frac{1}{2}x \right) dx$. Evaluating the integrals: $I = \left[ \frac{x^2}{4} - \frac{x^3}{3} \right]_{0}^{1/2} + \left[ \frac{x^3}{3} - \frac{x^2}{4} \right]_{1/2}^{1}$. $I = \left( \frac{(1/2)^2}{4} - \frac{(1/2)^3}{3} \right) + \left( \left( \frac{1}{3} - \frac{1}{4} \right) - \left( \frac{(1/2)^3}{3} - \frac{(1/2)^2}{4} \right) \right)$. $I = \left( \frac{1}{16} - \frac{1}{24} \right) + \left( \frac{1}{12} - \left( \frac{1}{24} - \frac{1}{16} \right) \right)$. $I = \left( \frac{3-2}{48} \right) + \left( \frac{1}{12} - \left( \frac{2-3}{48} \right) \right) = \frac{1}{48} + \frac{1}{12} + \frac{1}{48} = \frac{2}{48} + \frac{4}{48} = \frac{6}{48} = \frac{1}{8}$.

The value of $I = \int_{0}^{1} x \left| x - \frac{1}{2} \right| dx$ is

Similar Questions

The value of $\int_{-1 / \sqrt{2}}^{1 / \sqrt{2}}\left(\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2\right)^{1 / 2} d x$ is:

$\int_{1}^{5} (|x-3| + |1-x|) dx =$

If $f(x) = e^{-x} + 2e^{-2x} + 3e^{-3x} + \dots + \infty$,then $\int_{\ln 2}^{\ln 3} f(x) \, dx =$

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

The value of the definite integral $\int_{1}^{\infty} (e^{x+1} + e^{3-x})^{-1} \, dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module