What is the integral from 1 to 3 of sqrt(9-x^2) ?

The integral from 1 to 3 of sqrt(9-x^2) is (9pi)/4-sqrt(2)-9/2 arcsin(1/3)

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integral from 1 to 3 of sqrt(9-x^2)

\int_{1}^{3} 9 - x^{2} d x

\frac{9 π}{4} - 2 - \frac{9}{2} arcsin (\frac{1}{3})

Decimal Notation

4.12510 \dots

Solution steps

\int_{1}^{3} 9 - x^{2} d x

Apply Trigonometric Substitution

: \int_{a r c s i n (\frac{1}{3})}^{\frac{π}{2}} 9 cos^{2} (u) d u

= \int_{a r c s i n (\frac{1}{3})}^{\frac{π}{2}} 9 cos^{2} (u) d u

Take the constant out:

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

= 9 \cdot \int_{a r c s i n (\frac{1}{3})}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= 9 \cdot \int_{a r c s i n (\frac{1}{3})}^{\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

= 9 \cdot \frac{1}{2} \cdot \int_{a r c s i n (\frac{1}{3})}^{\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

= 9 \cdot \frac{1}{2} (\int_{a r c s i n (\frac{1}{3})}^{\frac{π}{2}} 1 d u + \int_{a r c s i n (\frac{1}{3})}^{\frac{π}{2}} cos (2 u) d u)

\int_{a r c s i n (\frac{1}{3})}^{\frac{π}{2}} 1 d u = \frac{π}{2} - arcsin (\frac{1}{3})

\int_{a r c s i n (\frac{1}{3})}^{\frac{π}{2}} cos (2 u) d u = - \frac{2 2}{9}

= 9 \cdot \frac{1}{2} (\frac{π}{2} - arcsin (\frac{1}{3}) - \frac{2 2}{9})

Simplify

9 \cdot \frac{1}{2} (\frac{π}{2} - arcsin (\frac{1}{3}) - \frac{2 2}{9}) : \frac{9 π}{4} - 2 - \frac{9}{2} arcsin (\frac{1}{3})

= \frac{9 π}{4} - 2 - \frac{9}{2} arcsin (\frac{1}{3})

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What is the integral from 1 to 3 of sqrt(9-x^2) ?
The integral from 1 to 3 of sqrt(9-x^2) is (9pi)/4-sqrt(2)-9/2 arcsin(1/3)