What is the integral from 0 to 9 of sqrt(162-x^2) ?

The integral from 0 to 9 of sqrt(162-x^2) is 81(pi/4+1/2)

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integral from 0 to 9 of sqrt(162-x^2)

\int_{0}^{9} 162 - x^{2} d x

81 (\frac{π}{4} + \frac{1}{2})

Decimal Notation

104.11725 \dots

Solution steps

\int_{0}^{9} 162 - x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{4}} 162 cos^{2} (u) d u

= \int_{0}^{\frac{π}{4}} 162 cos^{2} (u) d u

Take the constant out:

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

= 162 \cdot \int_{0}^{\frac{π}{4}} cos^{2} (u) d u

Rewrite using trig identities

= 162 \cdot \int_{0}^{\frac{π}{4}} \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

= 162 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} 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

= 162 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{4}} 1 d u + \int_{0}^{\frac{π}{4}} cos (2 u) d u)

\int_{0}^{\frac{π}{4}} 1 d u = \frac{π}{4}

\int_{0}^{\frac{π}{4}} cos (2 u) d u = \frac{1}{2}

= 162 \cdot \frac{1}{2} (\frac{π}{4} + \frac{1}{2})

Simplify

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

= 81 (\frac{π}{4} + \frac{1}{2})

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What is the integral from 0 to 9 of sqrt(162-x^2) ?
The integral from 0 to 9 of sqrt(162-x^2) is 81(pi/4+1/2)