What is the integral from-1 to 1 of sqrt(2-x^2)-x^2 ?

The integral from-1 to 1 of sqrt(2-x^2)-x^2 is 1/3+pi/2

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integral from-1 to 1 of sqrt(2-x^2)-x^2

\int_{- 1}^{1} 2 - x^{2} - x^{2} d x

\frac{1}{3} + \frac{π}{2}

Decimal Notation

1.90412 \dots

Solution steps

\int_{- 1}^{1} 2 - x^{2} - x^{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} 2 - x^{2} d x - \int_{- 1}^{1} x^{2} d x

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

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

= \frac{π}{2} + 1 - \frac{2}{3}

Simplify

\frac{π}{2} + 1 - \frac{2}{3} : \frac{1}{3} + \frac{π}{2}

= \frac{1}{3} + \frac{π}{2}

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