What is the integral from 0 to 1 of |x(2x-1)| ?

The integral from 0 to 1 of |x(2x-1)| is 1/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of |x(2x-1)|

\int_{0}^{1} ∣ x (2 x - 1) ∣ d x

\frac{1}{4}

Decimal

0.25

Solution steps

\int_{0}^{1} ∣ x (2 x - 1) ∣ d x

Eliminate Absolutes

= \int_{0}^{\frac{1}{2}} - 2 x^{2} + x d x + \int_{\frac{1}{2}}^{1} x (2 x - 1) d x

\int_{0}^{\frac{1}{2}} - 2 x^{2} + x d x = \frac{1}{24}

\int_{\frac{1}{2}}^{1} x (2 x - 1) d x = \frac{5}{24}

= \frac{1}{24} + \frac{5}{24}

Simplify

\frac{1}{24} + \frac{5}{24} : \frac{1}{4}

= \frac{1}{4}

\int_{0}^{3} e^{- t^{2}} d t

\int_{1}^{e} \frac{( 2 x ^{2} + 1 )}{2 x} d x

\int_{- 1}^{1} - x + 1 d x

\int_{0}^{π} (sec (x))^{2} d x

\int_{0}^{15} \frac{x}{( 1 + x ) ^{\frac{3}{4}}} d x

What is the integral from 0 to 1 of |x(2x-1)| ?
The integral from 0 to 1 of |x(2x-1)| is 1/4