What is the integral from 1 to 3 of (x^2+x+4) ?

The integral from 1 to 3 of (x^2+x+4) is 62/3

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integral from 1 to 3 of (x^2+x+4)

\int_{1}^{3} (x^{2} + x + 4) d x

\frac{62}{3}

Decimal Notation

20.66666 \dots

Solution steps

\int_{1}^{3} x^{2} + x + 4 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} 4 d x

\int_{1}^{3} x^{2} d x = \frac{26}{3}

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

\int_{1}^{3} 4 d x = 8

= \frac{26}{3} + 4 + 8

Simplify

\frac{26}{3} + 4 + 8 : \frac{62}{3}

= \frac{62}{3}

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What is the integral from 1 to 3 of (x^2+x+4) ?
The integral from 1 to 3 of (x^2+x+4) is 62/3