What is the integral from 9/2 to 9 of sqrt(81-x^2) ?

The integral from 9/2 to 9 of sqrt(81-x^2) is (27(4pi-3sqrt(3)))/8

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integral from 9/2 to 9 of sqrt(81-x^2)

\int_{\frac{9}{2}}^{9} 81 - x^{2} d x

\frac{27 ( 4 π - 3 3 )}{8}

Decimal

24.87448 \dots

Solution steps

\int_{\frac{9}{2}}^{9} 81 - x^{2} d x

Apply Trigonometric Substitution

: \int_{\frac{π}{6}}^{\frac{π}{2}} 81 cos^{2} (u) d u

= \int_{\frac{π}{6}}^{\frac{π}{2}} 81 cos^{2} (u) d u

Take the constant out:

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

= 81 \cdot \int_{\frac{π}{6}}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= 81 \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

= 81 \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

= 81 \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}

= 81 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4})

Simplify

81 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4}) : \frac{27 ( 4 π - 3 3 )}{8}

= \frac{27 ( 4 π - 3 3 )}{8}

x \to \infty lim ((1 - \frac{2}{x})^{x + 3})

\int 2 x^{\frac{2}{3}} d x

t^{2} y^{^{'}} + t y = t^{5}

\int \frac{y + 4}{y} d y

\int_{0}^{2} π (4 x)^{2} d x

What is the integral from 9/2 to 9 of sqrt(81-x^2) ?
The integral from 9/2 to 9 of sqrt(81-x^2) is (27(4pi-3sqrt(3)))/8