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

The integral from 0 to 1 of x^2sqrt(2-x^2) is pi/8

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integral from 0 to 1 of x^2sqrt(2-x^2)

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

\frac{π}{8}

Decimal Notation

0.39269 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{4}} 4 sin^{2} (u) cos^{2} (u) d u

= \int_{0}^{\frac{π}{4}} 4 sin^{2} (u) cos^{2} (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{0}^{\frac{π}{4}} sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= 4 \cdot \int_{0}^{\frac{π}{4}} \frac{1 - cos ( 4 u )}{8} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \frac{1}{8} \cdot \int_{0}^{\frac{π}{4}} 1 - cos (4 u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 \cdot \frac{1}{8} (\int_{0}^{\frac{π}{4}} 1 d u - \int_{0}^{\frac{π}{4}} cos (4 u) d u)

\int_{0}^{\frac{π}{4}} 1 d u = \frac{π}{4}

\int_{0}^{\frac{π}{4}} cos (4 u) d u = 0

= 4 \cdot \frac{1}{8} (\frac{π}{4} - 0)

Simplify

4 \cdot \frac{1}{8} (\frac{π}{4} - 0) : \frac{π}{8}

= \frac{π}{8}

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What is the integral from 0 to 1 of x^2sqrt(2-x^2) ?
The integral from 0 to 1 of x^2sqrt(2-x^2) is pi/8