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

The integral from-1 to 1 of x^2sqrt(4-x^2) is 2(pi/3-(sqrt(3))/4)

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integral from-1 to 1 of x^2sqrt(4-x^2)

\int_{- 1}^{1} x^{2} 4 - x^{2} d x

2 (\frac{π}{3} - \frac{3}{4})

Decimal

1.22836 \dots

Solution steps

\int_{- 1}^{1} x^{2} 4 - x^{2} d x

Apply Trigonometric Substitution

= \int_{- \frac{π}{6}}^{\frac{π}{6}} 16 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

= 16 \cdot \int_{- \frac{π}{6}}^{\frac{π}{6}} sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= 16 \cdot \int_{- \frac{π}{6}}^{\frac{π}{6}} \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

= 16 \cdot \frac{1}{8} \cdot \int_{- \frac{π}{6}}^{\frac{π}{6}} 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

= 16 \cdot \frac{1}{8} (\int_{- \frac{π}{6}}^{\frac{π}{6}} 1 d u - \int_{- \frac{π}{6}}^{\frac{π}{6}} cos (4 u) d u)

\int_{- \frac{π}{6}}^{\frac{π}{6}} 1 d u = \frac{π}{3}

\int_{- \frac{π}{6}}^{\frac{π}{6}} cos (4 u) d u = \frac{3}{4}

= 16 \cdot \frac{1}{8} (\frac{π}{3} - \frac{3}{4})

Simplify

16 \cdot \frac{1}{8} (\frac{π}{3} - \frac{3}{4}) : 2 (\frac{π}{3} - \frac{3}{4})

= 2 (\frac{π}{3} - \frac{3}{4})

What is the integral from-1 to 1 of x^2sqrt(4-x^2) ?
The integral from-1 to 1 of x^2sqrt(4-x^2) is 2(pi/3-(sqrt(3))/4)