What is the integral from 0 to 8 of x^2sqrt(64-x^2) ?

The integral from 0 to 8 of x^2sqrt(64-x^2) is 256pi

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integral from 0 to 8 of x^2sqrt(64-x^2)

\int_{0}^{8} x^{2} 64 - x^{2} d x

256 π

Decimal Notation

804.24771 \dots

Solution steps

\int_{0}^{8} x^{2} 64 - x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 4096 sin^{2} (u) cos^{2} (u) d u

= \int_{0}^{\frac{π}{2}} 4096 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

= 4096 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= 4096 \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

= 4096 \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

= 4096 \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

= 4096 \cdot \frac{1}{8} (\frac{π}{2} - 0)

Simplify

4096 \cdot \frac{1}{8} (\frac{π}{2} - 0) : 256 π

= 256 π

y^{^{'}} - 5 y = 6 e^{7 t}

5 x - x^{2}

\int (- sin (x) + \frac{1}{cos ^{2} ( x )}) d x

\int \frac{e ^{2 x}}{81 + e ^{4 x}} d x

e^{2 x}, e^{x}, x = ln (3)

What is the integral from 0 to 8 of x^2sqrt(64-x^2) ?
The integral from 0 to 8 of x^2sqrt(64-x^2) is 256pi