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Popular >

integral from 0 to 4 of |2x-3|

Solution

\int_{0}^{4} ∣ 2 x - 3 ∣ d x

Solution

\frac{17}{2}

+1

Solution steps

\int_{0}^{4} ∣ 2 x - 3 ∣ d x

Eliminate Absolutes

= \int_{0}^{\frac{3}{2}} - 2 x + 3 d x + \int_{\frac{3}{2}}^{4} 2 x - 3 d x

\int_{0}^{\frac{3}{2}} - 2 x + 3 d x = \frac{9}{4}

\int_{\frac{3}{2}}^{4} 2 x - 3 d x = \frac{25}{4}

= \frac{9}{4} + \frac{25}{4}

Simplify

\frac{9}{4} + \frac{25}{4} : \frac{17}{2}

= \frac{17}{2}

Popular Examples

inverse oflaplace s/(s+1/2)inverse laplace

\frac{s}{s + \frac{1}{2}}

maclaurin e^{-x},5 maclaurin

e^{- x}, 5

integral of 1+2/y

\int 1 + \frac{2}{y} d y

derivative of 1/(2z^3)derivative of

\frac{1}{2 z ^{3}}

y^'=-(2y)/(x+1)+2x

y^{^{'}} = - \frac{2 y}{x + 1} + 2 x

Frequently Asked Questions (FAQ)

What is the integral from 0 to 4 of |2x-3| ?
The integral from 0 to 4 of |2x-3| is 17/2