What is the integral from 0 to 1 of sqrt(9-x^2) ?

The integral from 0 to 1 of sqrt(9-x^2) is 9/2 arcsin(1/3)+sqrt(2)

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integral from 0 to 1 of sqrt(9-x^2)

\int_{0}^{1} 9 - x^{2} d x

\frac{9}{2} arcsin (\frac{1}{3}) + 2

Decimal

2.94347 \dots

Solution steps

\int_{0}^{1} 9 - x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c s i n (\frac{1}{3})} 9 cos^{2} (u) d u

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

Rewrite using trig identities

= 9 \cdot \int_{0}^{a r c s i n (\frac{1}{3})} \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_{0}^{a r c s i n (\frac{1}{3})} 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_{0}^{a r c s i n (\frac{1}{3})} 1 d u + \int_{0}^{a r c s i n (\frac{1}{3})} cos (2 u) d u)

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

\int_{0}^{a r c s i n (\frac{1}{3})} cos (2 u) d u = \frac{2 2}{9}

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

Simplify

9 \cdot \frac{1}{2} (arcsin (\frac{1}{3}) + \frac{2 2}{9}) : \frac{9}{2} arcsin (\frac{1}{3}) + 2

= \frac{9}{2} arcsin (\frac{1}{3}) + 2

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What is the integral from 0 to 1 of sqrt(9-x^2) ?
The integral from 0 to 1 of sqrt(9-x^2) is 9/2 arcsin(1/3)+sqrt(2)