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

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

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

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

\frac{π}{8}

Decimal Notation

0.39269 \dots

Solution steps

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= \int_{- \frac{π}{2}}^{\frac{π}{2}} \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

= \frac{1}{8} \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} 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

= \frac{1}{8} (\int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d u - \int_{- \frac{π}{2}}^{\frac{π}{2}} cos (4 u) d u)

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

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

= \frac{1}{8} (π - 0)

Simplify

\frac{1}{8} (π - 0) : \frac{π}{8}

= \frac{π}{8}

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

\int_{0}^{\frac{π}{2}} cos (x) + sin (x) d x

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

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

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

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