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Popular >

integral from 1 to 4 of (x^2+x)

Solution

\int_{1}^{4} (x^{2} + x) d x

Solution

\frac{57}{2}

+1

Decimal

28.5

Solution steps

\int_{1}^{4} 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}^{4} x^{2} d x + \int_{1}^{4} x d x

\int_{1}^{4} x^{2} d x = 21

\int_{1}^{4} x d x = \frac{15}{2}

= 21 + \frac{15}{2}

Simplify

21 + \frac{15}{2} : \frac{57}{2}

= \frac{57}{2}

Graph

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Frequently Asked Questions (FAQ)

What is the integral from 1 to 4 of (x^2+x) ?
The integral from 1 to 4 of (x^2+x) is 57/2