What is the integral from 0 to 1 of rsqrt(2-r^2)-r^2 ?

The integral from 0 to 1 of rsqrt(2-r^2)-r^2 is (2sqrt(2)-1)/3-1/3

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integral from 0 to 1 of rsqrt(2-r^2)-r^2

\int_{0}^{1} r 2 - r^{2} - r^{2} d r

\frac{2 2 - 1}{3} - \frac{1}{3}

Decimal Notation

0.27614 \dots

Solution steps

\int_{0}^{1} r 2 - r^{2} - r^{2} d r

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} r 2 - r^{2} d r - \int_{0}^{1} r^{2} d r

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

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

= \frac{2 2 - 1}{3} - \frac{1}{3}

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

\int_{- 1}^{1} \frac{x}{cos ^{2} ( x )} d x

\int_{0}^{3} (4 x^{2} - 2 x + 5) d x

\int_{0}^{2} 5 x e^{- x^{2}} d x

\int_{0}^{1} 3 x (4 x^{2} - 1)^{5} d x

What is the integral from 0 to 1 of rsqrt(2-r^2)-r^2 ?
The integral from 0 to 1 of rsqrt(2-r^2)-r^2 is (2sqrt(2)-1)/3-1/3