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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

16 π

Decimal

50.26548 \dots

Solution steps

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

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} x^{3} - x d x

\int_{1}^{- 1} (r + 1)^{2} d r

\int_{\frac{π}{4}}^{\frac{π}{2}} 2 - csc^{2} (x) d x

\int_{1}^{3} x^{3} - 4 x^{2} + 3 x d x

\int_{x}^{4 x} 2^{t} d t

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