What is the integral from 0 to 3 of x^2sqrt(9-x^2) ?

The integral from 0 to 3 of x^2sqrt(9-x^2) is (81pi)/(16)

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integral from 0 to 3 of x^2sqrt(9-x^2)

\int_{0}^{3} x^{2} 9 - x^{2} d x

\frac{81 π}{16}

Decimal Notation

15.90431 \dots

Solution steps

\int_{0}^{3} x^{2} 9 - x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 81 sin^{2} (u) cos^{2} (u) d u

= \int_{0}^{\frac{π}{2}} 81 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

= 81 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= 81 \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

= 81 \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

= 81 \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

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

Simplify

81 \cdot \frac{1}{8} (\frac{π}{2} - 0) : \frac{81 π}{16}

= \frac{81 π}{16}

\int \frac{1}{t ^{2} 256 - t ^{2}} d t

n = 0 \sum \infty (\frac{n x}{6})^{n}

x \to + 2 lim (2 - x)

f (x) = (2 t - 1) (3 t - 4)^{- 1}

2^{4 x} + 4 x^{2}

What is the integral from 0 to 3 of x^2sqrt(9-x^2) ?
The integral from 0 to 3 of x^2sqrt(9-x^2) is (81pi)/(16)