What is the integral from 0 to 1 of 1-x^2+(x^4)/2 ?

The integral from 0 to 1 of 1-x^2+(x^4)/2 is 23/30

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integral from 0 to 1 of 1-x^2+(x^4)/2

\int_{0}^{1} 1 - x^{2} + \frac{x ^{4}}{2} d x

\frac{23}{30}

Decimal

0.76666 \dots

Solution steps

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

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

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

\int_{0}^{1} \frac{x ^{4}}{2} d x = \frac{1}{10}

= 1 - \frac{1}{3} + \frac{1}{10}

Simplify

1 - \frac{1}{3} + \frac{1}{10} : \frac{23}{30}

= \frac{23}{30}

What is the integral from 0 to 1 of 1-x^2+(x^4)/2 ?
The integral from 0 to 1 of 1-x^2+(x^4)/2 is 23/30