What is the integral from 9 to 18 of sqrt(324-x^2) ?

The integral from 9 to 18 of sqrt(324-x^2) is 162(pi/3-(sqrt(3))/4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 9 to 18 of sqrt(324-x^2)

\int_{9}^{18} 324 - x^{2} d x

162 (\frac{π}{3} - \frac{3}{4})

Decimal Notation

99.49794 \dots

Solution steps

\int_{9}^{18} 324 - x^{2} d x

Apply Trigonometric Substitution

: \int_{\frac{π}{6}}^{\frac{π}{2}} 324 cos^{2} (u) d u

= \int_{\frac{π}{6}}^{\frac{π}{2}} 324 cos^{2} (u) d u

Take the constant out:

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

= 324 \cdot \int_{\frac{π}{6}}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= 324 \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

= 324 \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

= 324 \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}

= 324 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4})

Simplify

324 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4}) : 162 (\frac{π}{3} - \frac{3}{4})

= 162 (\frac{π}{3} - \frac{3}{4})

\int cot (11 x) d x

\frac{d}{d x} (\frac{2}{1 + 4 x})

(16, - 2), (15, 86)

- \frac{200}{( t + 2 ) ^{2}}

\frac{d}{d x} (\frac{3}{x} + 5 sin (x))

What is the integral from 9 to 18 of sqrt(324-x^2) ?
The integral from 9 to 18 of sqrt(324-x^2) is 162(pi/3-(sqrt(3))/4)