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

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

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

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

9 arcsin (\frac{2}{3}) + 25

Decimal Notation

11.03968 \dots

Solution steps

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

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

\int_{- a r c s i n (\frac{2}{3})}^{a r c s i n (\frac{2}{3})} cos (2 u) d u = \frac{4 5}{9}

= 9 \cdot \frac{1}{2} (2 arcsin (\frac{2}{3}) + \frac{4 5}{9})

Simplify

9 \cdot \frac{1}{2} (2 arcsin (\frac{2}{3}) + \frac{4 5}{9}) : 9 arcsin (\frac{2}{3}) + 25

= 9 arcsin (\frac{2}{3}) + 25

\int_{0.5}^{0.9} 0.0014400000000 \dots x d x

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