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

The integral from 3/2 to 3 of sqrt(9-x^2) is (3(4pi-3sqrt(3)))/8

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integral from 3/2 to 3 of sqrt(9-x^2)

\int_{\frac{3}{2}}^{3} 9 - x^{2} d x

\frac{3 ( 4 π - 3 3 )}{8}

Decimal Notation

2.76383 \dots

Solution steps

\int_{\frac{3}{2}}^{3} 9 - x^{2} d x

Apply Trigonometric Substitution

: \int_{\frac{π}{6}}^{\frac{π}{2}} 9 cos^{2} (u) d u

= \int_{\frac{π}{6}}^{\frac{π}{2}} 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_{\frac{π}{6}}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= 9 \cdot \int_{\frac{π}{6}}^{\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

= 9 \cdot \frac{1}{2} \cdot \int_{\frac{π}{6}}^{\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

= 9 \cdot \frac{1}{2} (\int_{\frac{π}{6}}^{\frac{π}{2}} 1 d u + \int_{\frac{π}{6}}^{\frac{π}{2}} cos (2 u) d u)

\int_{\frac{π}{6}}^{\frac{π}{2}} 1 d u = \frac{π}{3}

\int_{\frac{π}{6}}^{\frac{π}{2}} cos (2 u) d u = - \frac{3}{4}

= 9 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4})

Simplify

9 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4}) : \frac{3 ( 4 π - 3 3 )}{8}

= \frac{3 ( 4 π - 3 3 )}{8}

\frac{d}{d x} (cos (x^{2}))

\int_{0}^{3} \frac{1}{x ^{2} - x - 2} d x

x \to 3 lim (3 x^{2} - 2 x + 5)

\int \frac{sin ( x )}{cos ( x )} d x

\int 2 (3 x + 4)^{- 4} d x

What is the integral from 3/2 to 3 of sqrt(9-x^2) ?
The integral from 3/2 to 3 of sqrt(9-x^2) is (3(4pi-3sqrt(3)))/8