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Popular Calculus >

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

Graph

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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