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

The integral from 0 to 2 of sqrt(9-x^2) is 9/2 arcsin(2/3)+sqrt(5)

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

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

\frac{9}{2} arcsin (\frac{2}{3}) + 5

Decimal Notation

5.51984 \dots

Solution steps

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

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

\int_{0}^{a r c s i n (\frac{2}{3})} cos (2 u) d u = \frac{2 5}{9}

= 9 \cdot \frac{1}{2} (arcsin (\frac{2}{3}) + \frac{2 5}{9})

Simplify

9 \cdot \frac{1}{2} (arcsin (\frac{2}{3}) + \frac{2 5}{9}) : \frac{9}{2} arcsin (\frac{2}{3}) + 5

= \frac{9}{2} arcsin (\frac{2}{3}) + 5

g (x) = \frac{5 + 2 x}{7 - 8 x}

\int sin (k t) d t

- t^{2} + 9 t - 20

\int \frac{6 x ^{2}}{( 2 x - 1 ) ( 2 x + 3 )} d x

(x^{2} - 3 x y) d x + x d y = 0

What is the integral from 0 to 2 of sqrt(9-x^2) ?
The integral from 0 to 2 of sqrt(9-x^2) is 9/2 arcsin(2/3)+sqrt(5)