What is the integral from 0 to 9 of x^2sqrt(81-x^2) ?

The integral from 0 to 9 of x^2sqrt(81-x^2) is (6561pi)/(16)

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integral from 0 to 9 of x^2sqrt(81-x^2)

\int_{0}^{9} x^{2} 81 - x^{2} d x

\frac{6561 π}{16}

Decimal Notation

1288.24933 \dots

Solution steps

\int_{0}^{9} x^{2} 81 - x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 6561 sin^{2} (u) cos^{2} (u) d u

= \int_{0}^{\frac{π}{2}} 6561 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

= 6561 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= 6561 \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

= 6561 \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

= 6561 \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

= 6561 \cdot \frac{1}{8} (\frac{π}{2} - 0)

Simplify

6561 \cdot \frac{1}{8} (\frac{π}{2} - 0) : \frac{6561 π}{16}

= \frac{6561 π}{16}

\int_{0.1}^{x} x d x

0, 5 x, x^{2} - 6

\int 100 x^{\frac{3}{2}} d x

\frac{\partial}{\partial x} (\frac{4 π ^{2} y}{x ^{2}})

\frac{d}{d x} (\frac{1}{a x ^{2} + b x + c})

What is the integral from 0 to 9 of x^2sqrt(81-x^2) ?
The integral from 0 to 9 of x^2sqrt(81-x^2) is (6561pi)/(16)