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Popular >

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

Solution

\int_{- 1}^{2} (2 + ∣ x ∣) d x

Solution

\frac{17}{2}

+1

Decimal

8.5

Solution steps

\int_{- 1}^{2} 2 + ∣ x ∣ d x

Eliminate Absolutes

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

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

\int_{0}^{2} 2 + x d x = 6

= \frac{5}{2} + 6

Simplify

\frac{5}{2} + 6 : \frac{17}{2}

= \frac{17}{2}

Graph

Popular Examples

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\int_{0}^{2} (\frac{1}{2} x + \frac{1}{2}) - x^{2} d x

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\int_{0}^{1} (\frac{1}{3 x} - 2 x + 1) d x

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

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\int_{0}^{2 π} \frac{π}{4} θ^{2} d θ

integral from 1 to 2 of (-x^2+3x-2)

\int_{1}^{2} (- x^{2} + 3 x - 2) d x

Frequently Asked Questions (FAQ)

What is the integral from-1 to 2 of (2+|x|) ?
The integral from-1 to 2 of (2+|x|) is 17/2