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Popular >

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

Solution

\int_{- 1}^{2} x + ∣ x ∣ d x

Solution

4

Solution steps

\int_{- 1}^{2} x + ∣ x ∣ d x

Eliminate Absolutes

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

\int_{- 1}^{0} x - x d x = 0

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

= 0 + 4

Simplify

= 4

Graph

Popular Examples

integral from-pi to 0 of (x+pi)cos(nx)

\int_{- π}^{0} (x + π) cos (n x) d x

integral from 0 to x/2 of (sin(x))/x

\int_{0}^{\frac{x}{2}} \frac{sin ( x )}{x} d x

integral from-2 to 2 of 2x^3

\int_{- 2}^{2} 2 x^{3} d x

integral from 0 to 1 of x^4e^{x^5}

\int_{0}^{1} x^{4} e^{x^{5}} d x

integral from 0 to 4 of 1/(16x^2+9)

\int_{0}^{4} \frac{1}{16 x ^{2} + 9} d x

Frequently Asked Questions (FAQ)

What is the integral from-1 to 2 of x+|x| ?
The integral from-1 to 2 of x+|x| is 4