What is the integral from 0 to 3 of sqrt(25-x^2) ?

The integral from 0 to 3 of sqrt(25-x^2) is 25/2 arcsin(3/5)+6

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integral from 0 to 3 of sqrt(25-x^2)

\int_{0}^{3} 25 - x^{2} d x

\frac{25}{2} arcsin (\frac{3}{5}) + 6

Solution steps

\int_{0}^{3} 25 - x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c s i n (\frac{3}{5})} 25 cos^{2} (u) d u

= \int_{0}^{a r c s i n (\frac{3}{5})} 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_{0}^{a r c s i n (\frac{3}{5})} cos^{2} (u) d u

Rewrite using trig identities

= 25 \cdot \int_{0}^{a r c s i n (\frac{3}{5})} \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_{0}^{a r c s i n (\frac{3}{5})} 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_{0}^{a r c s i n (\frac{3}{5})} 1 d u + \int_{0}^{a r c s i n (\frac{3}{5})} cos (2 u) d u)

\int_{0}^{a r c s i n (\frac{3}{5})} 1 d u = arcsin (\frac{3}{5})

\int_{0}^{a r c s i n (\frac{3}{5})} cos (2 u) d u = \frac{12}{25}

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

Simplify

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

= \frac{25}{2} arcsin (\frac{3}{5}) + 6

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What is the integral from 0 to 3 of sqrt(25-x^2) ?
The integral from 0 to 3 of sqrt(25-x^2) is 25/2 arcsin(3/5)+6