What is the integral from 8 to 16 of sqrt(256-x^2) ?

The integral from 8 to 16 of sqrt(256-x^2) is 128(pi/3-(sqrt(3))/4)

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integral from 8 to 16 of sqrt(256-x^2)

\int_{8}^{16} 256 - x^{2} d x

128 (\frac{π}{3} - \frac{3}{4})

Decimal Notation

78.61566 \dots

Solution steps

\int_{8}^{16} 256 - x^{2} d x

Apply Trigonometric Substitution

: \int_{\frac{π}{6}}^{\frac{π}{2}} 256 cos^{2} (u) d u

= \int_{\frac{π}{6}}^{\frac{π}{2}} 256 cos^{2} (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 256 \cdot \int_{\frac{π}{6}}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= 256 \cdot \int_{\frac{π}{6}}^{\frac{π}{2}} \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 256 \cdot \frac{1}{2} \cdot \int_{\frac{π}{6}}^{\frac{π}{2}} 1 + cos (2 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

= 256 \cdot \frac{1}{2} (\int_{\frac{π}{6}}^{\frac{π}{2}} 1 d u + \int_{\frac{π}{6}}^{\frac{π}{2}} cos (2 u) d u)

\int_{\frac{π}{6}}^{\frac{π}{2}} 1 d u = \frac{π}{3}

\int_{\frac{π}{6}}^{\frac{π}{2}} cos (2 u) d u = - \frac{3}{4}

= 256 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4})

Simplify

256 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4}) : 128 (\frac{π}{3} - \frac{3}{4})

= 128 (\frac{π}{3} - \frac{3}{4})

\int_{- π}^{π} sin (2 x) d x

\frac{d}{d x} (\frac{3 x ^{2} + x}{- x ^{2} + 3 x - 3})

\frac{\partial}{\partial x} (\frac{2 x y}{x - y})

f (x) = \frac{x}{7}

y = - x^{2} - 8 x - 12, y = 0, [- 4, - 1]

What is the integral from 8 to 16 of sqrt(256-x^2) ?
The integral from 8 to 16 of sqrt(256-x^2) is 128(pi/3-(sqrt(3))/4)