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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

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\int_{0}^{2} (6 - t) t d t

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\int_{0}^{\infty} e^{- n x} d x

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\int_{1}^{N} \frac{1}{x ^{p}} d x

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\int_{1}^{3} \frac{7}{x} d x

integral from 1 to t of xln(x)

\int_{1}^{t} x ln (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