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

The integral from-1 to 1 of (1/3+(x^4)/2) is 13/15

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

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

\frac{13}{15}

Decimal Notation

0.86666 \dots

Solution steps

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

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

\int_{- 1}^{1} \frac{x ^{4}}{2} d x = \frac{1}{5}

= \frac{2}{3} + \frac{1}{5}

Simplify

\frac{2}{3} + \frac{1}{5} : \frac{13}{15}

= \frac{13}{15}

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What is the integral from-1 to 1 of (1/3+(x^4)/2) ?
The integral from-1 to 1 of (1/3+(x^4)/2) is 13/15