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

The integral from 0 to 1 of (x^2+2)^2 is 83/15

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

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

\frac{83}{15}

Decimal

5.53333 \dots

Solution steps

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

Expand

(x^{2} + 2)^{2} : x^{4} + 4 x^{2} + 4

= \int_{0}^{1} x^{4} + 4 x^{2} + 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_{0}^{1} x^{4} d x + \int_{0}^{1} 4 x^{2} d x + \int_{0}^{1} 4 d x

\int_{0}^{1} x^{4} d x = \frac{1}{5}

\int_{0}^{1} 4 x^{2} d x = \frac{4}{3}

\int_{0}^{1} 4 d x = 4

= \frac{1}{5} + \frac{4}{3} + 4

Simplify

\frac{1}{5} + \frac{4}{3} + 4 : \frac{83}{15}

= \frac{83}{15}

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What is the integral from 0 to 1 of (x^2+2)^2 ?
The integral from 0 to 1 of (x^2+2)^2 is 83/15