What is the integral from 4 to 9 of (sqrt(x)+x) ?

The integral from 4 to 9 of (sqrt(x)+x) is 271/6

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integral from 4 to 9 of (sqrt(x)+x)

\int_{4}^{9} (x + x) d x

\frac{271}{6}

Decimal

45.16666 \dots

Solution steps

\int_{4}^{9} x + 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_{4}^{9} x d x + \int_{4}^{9} x d x

\int_{4}^{9} x d x = \frac{38}{3}

\int_{4}^{9} x d x = \frac{65}{2}

= \frac{38}{3} + \frac{65}{2}

Simplify

\frac{38}{3} + \frac{65}{2} : \frac{271}{6}

= \frac{271}{6}

What is the integral from 4 to 9 of (sqrt(x)+x) ?
The integral from 4 to 9 of (sqrt(x)+x) is 271/6