What is the integral from 0 to 9 of sqrt(81-x^2) ?

The integral from 0 to 9 of sqrt(81-x^2) is (81pi)/4

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integral from 0 to 9 of sqrt(81-x^2)

\int_{0}^{9} 81 - x^{2} d x

\frac{81 π}{4}

Decimal Notation

63.61725 \dots

Solution steps

\int_{0}^{9} 81 - x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 81 cos^{2} (u) d u

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

Rewrite using trig identities

= 81 \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

= 81 \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

= 81 \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

= 81 \cdot \frac{1}{2} (\frac{π}{2} + 0)

Simplify

81 \cdot \frac{1}{2} (\frac{π}{2} + 0) : \frac{81 π}{4}

= \frac{81 π}{4}

\int \frac{1}{x ^{2} 2 x ^{2} + 32} d x

64 y^{^{''}} + 144 y^{^{'}} + 81 y = 0

\frac{10}{s ^{2} + s + 10} \cdot \frac{1}{s}

\frac{\partial}{\partial x} (\frac{e ^{y}}{x + y ^{5}})

\frac{\partial}{\partial x} (cos (2 c x + a c t))

What is the integral from 0 to 9 of sqrt(81-x^2) ?
The integral from 0 to 9 of sqrt(81-x^2) is (81pi)/4