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

The integral from 0 to 8 of sqrt(64-x^2) is 16pi

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

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

16 π

Decimal Notation

50.26548 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 64 cos^{2} (u) d u

= \int_{0}^{\frac{π}{2}} 64 cos^{2} (u) d u

Take the constant out:

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

= 64 \cdot \int_{0}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= 64 \cdot \int_{0}^{\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

= 64 \cdot \frac{1}{2} \cdot \int_{0}^{\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

= 64 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d u + \int_{0}^{\frac{π}{2}} cos (2 u) d u)

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

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

= 64 \cdot \frac{1}{2} (\frac{π}{2} + 0)

Simplify

64 \cdot \frac{1}{2} (\frac{π}{2} + 0) : 16 π

= 16 π

\int_{0}^{1} t e^{- 1} d t

(x^{2} + 1) (\frac{d y}{d x}) + 8 x (y - 1) = 0, y (0) = 5

\int 3 x^{5} e^{x} d x

6 x + \frac{d y}{d x} = 0

\frac{\partial}{\partial x} (x (x^{2} + y^{2} + z^{2})^{m})

What is the integral from 0 to 8 of sqrt(64-x^2) ?
The integral from 0 to 8 of sqrt(64-x^2) is 16pi