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Popular Calculus >

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

Solution

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

Solution

\frac{40}{3}

+1

Decimal

13.33333 \dots

Solution steps

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

\int_{- 1}^{3} x^{2} d x = \frac{28}{3}

\int_{- 1}^{3} x d x = 4

= \frac{28}{3} + 4

Simplify

\frac{28}{3} + 4 : \frac{40}{3}

= \frac{40}{3}

Graph

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Frequently Asked Questions (FAQ)

What is the integral from-1 to 3 of (x^2+x) ?
The integral from-1 to 3 of (x^2+x) is 40/3