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

The integral from-8 to 8 of (sqrt(64-x^2)) is 32pi

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

\int_{- 8}^{8} (64 - x^{2}) d x

32 π

Decimal Notation

100.53096 \dots

Solution steps

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

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

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

= 64 \cdot \frac{1}{2} (π + 0)

Simplify

64 \cdot \frac{1}{2} (π + 0) : 32 π

= 32 π

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What is the integral from-8 to 8 of (sqrt(64-x^2)) ?
The integral from-8 to 8 of (sqrt(64-x^2)) is 32pi