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Popular >

integral from 0 to 5 of 1+|x-2|

Solution

\int_{0}^{5} 1 + ∣ x - 2 ∣ d x

Solution

\frac{23}{2}

+1

Decimal

11.5

Solution steps

\int_{0}^{5} 1 + ∣ x - 2 ∣ d x

Eliminate Absolutes

= \int_{0}^{2} 1 - x + 2 d x + \int_{2}^{5} 1 + x - 2 d x

\int_{0}^{2} 1 - x + 2 d x = 4

\int_{2}^{5} 1 + x - 2 d x = \frac{15}{2}

= 4 + \frac{15}{2}

Simplify

4 + \frac{15}{2} : \frac{23}{2}

= \frac{23}{2}

Graph

Popular Examples

integral from 1 to 13 of 2x+3

\int_{1}^{13} 2 x + 3 d x

integral from 1 to e^2 of (ln^2(x))/x

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

integral from-1 to 0 of-3-x^3+3-x

\int_{- 1}^{0} - 3 - x^{3} + 3 - x d x

integral from 0 to 1 of x^2+x+2

\int_{0}^{1} x^{2} + x + 2 d x

integral from 1 to 7 of (1/(x^2))

\int_{1}^{7} (\frac{1}{x ^{2}}) d x

Frequently Asked Questions (FAQ)

What is the integral from 0 to 5 of 1+|x-2| ?
The integral from 0 to 5 of 1+|x-2| is 23/2