What is the integral from 0 to 6 of x^2sqrt(36-x^2) ?

The integral from 0 to 6 of x^2sqrt(36-x^2) is 81pi

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integral from 0 to 6 of x^2sqrt(36-x^2)

\int_{0}^{6} x^{2} 36 - x^{2} d x

81 π

Decimal

254.46900 \dots

Solution steps

\int_{0}^{6} x^{2} 36 - x^{2} d x

Apply Trigonometric Substitution

= \int_{0}^{\frac{π}{2}} 1296 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

= 1296 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= 1296 \cdot \int_{0}^{\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

= 1296 \cdot \frac{1}{8} \cdot \int_{0}^{\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

= 1296 \cdot \frac{1}{8} (\int_{0}^{\frac{π}{2}} 1 d u - \int_{0}^{\frac{π}{2}} cos (4 u) d u)

\int_{0}^{\frac{π}{2}} 1 d u = \frac{π}{2}

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

= 1296 \cdot \frac{1}{8} (\frac{π}{2} - 0)

Simplify

1296 \cdot \frac{1}{8} (\frac{π}{2} - 0) : 81 π

= 81 π

What is the integral from 0 to 6 of x^2sqrt(36-x^2) ?
The integral from 0 to 6 of x^2sqrt(36-x^2) is 81pi