What is the integral from 3 to 5 of sqrt(25-x^2) ?

The integral from 3 to 5 of sqrt(25-x^2) is (25pi)/4-6-25/2 arcsin(3/5)

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integral from 3 to 5 of sqrt(25-x^2)

\int_{3}^{5} 25 - x^{2} d x

\frac{25 π}{4} - 6 - \frac{25}{2} arcsin (\frac{3}{5})

Decimal Notation

5.59119 \dots

Solution steps

\int_{3}^{5} 25 - x^{2} d x

Apply Trigonometric Substitution

: \int_{a r c s i n (\frac{3}{5})}^{\frac{π}{2}} 25 cos^{2} (u) d u

= \int_{a r c s i n (\frac{3}{5})}^{\frac{π}{2}} 25 cos^{2} (u) d u

Take the constant out:

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

= 25 \cdot \int_{a r c s i n (\frac{3}{5})}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= 25 \cdot \int_{a r c s i n (\frac{3}{5})}^{\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

= 25 \cdot \frac{1}{2} \cdot \int_{a r c s i n (\frac{3}{5})}^{\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

= 25 \cdot \frac{1}{2} (\int_{a r c s i n (\frac{3}{5})}^{\frac{π}{2}} 1 d u + \int_{a r c s i n (\frac{3}{5})}^{\frac{π}{2}} cos (2 u) d u)

\int_{a r c s i n (\frac{3}{5})}^{\frac{π}{2}} 1 d u = \frac{π}{2} - arcsin (\frac{3}{5})

\int_{a r c s i n (\frac{3}{5})}^{\frac{π}{2}} cos (2 u) d u = - \frac{12}{25}

= 25 \cdot \frac{1}{2} (\frac{π}{2} - arcsin (\frac{3}{5}) - \frac{12}{25})

Simplify

25 \cdot \frac{1}{2} (\frac{π}{2} - arcsin (\frac{3}{5}) - \frac{12}{25}) : \frac{25 π}{4} - 6 - \frac{25}{2} arcsin (\frac{3}{5})

= \frac{25 π}{4} - 6 - \frac{25}{2} arcsin (\frac{3}{5})

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What is the integral from 3 to 5 of sqrt(25-x^2) ?
The integral from 3 to 5 of sqrt(25-x^2) is (25pi)/4-6-25/2 arcsin(3/5)