What is the integral from 0 to 5 of ((x^2)/3+1) ?

The integral from 0 to 5 of ((x^2)/3+1) is 170/9

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integral from 0 to 5 of ((x^2)/3+1)

\int_{0}^{5} (\frac{x ^{2}}{3} + 1) d x

\frac{170}{9}

Decimal

18.88888 \dots

Solution steps

\int_{0}^{5} \frac{x ^{2}}{3} + 1 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_{0}^{5} \frac{x ^{2}}{3} d x + \int_{0}^{5} 1 d x

\int_{0}^{5} \frac{x ^{2}}{3} d x = \frac{125}{9}

\int_{0}^{5} 1 d x = 5

= \frac{125}{9} + 5

Simplify

\frac{125}{9} + 5 : \frac{170}{9}

= \frac{170}{9}

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What is the integral from 0 to 5 of ((x^2)/3+1) ?
The integral from 0 to 5 of ((x^2)/3+1) is 170/9