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Popular >

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

Solution

\int_{- 1}^{1} ∣ x + 2 ∣ d x

Solution

4

Solution steps

\int_{- 1}^{1} ∣ x + 2 ∣ d x

Eliminate Absolutes

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

= 0 + 4

Simplify

= 4

Popular Examples

integral from-2021 to 2021 of x^5

\int_{- 2021}^{2021} x^{5} d x

integral from 1 to 2 of xsqrt(2x-1)

\int_{1}^{2} x 2 x - 1 d x

integral from 1 to x of 2x^2y^{-2}+2y

\int_{1}^{x} 2 x^{2} y^{- 2} + 2 y d y

integral from 0 to 5 of 3x^2

\int_{0}^{5} 3 x^{2} d x

integral from 0 to pi/2 of (8+4cos(x))

\int_{0}^{\frac{π}{2}} (8 + 4 cos (x)) d x

Frequently Asked Questions (FAQ)

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