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

The integral from 0 to 3/2 of sqrt(9-x^2) is (3(2pi+3sqrt(3)))/8

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integral from 0 to 3/2 of sqrt(9-x^2)

\int_{0}^{\frac{3}{2}} 9 - x^{2} d x

\frac{3 ( 2 π + 3 3 )}{8}

Decimal Notation

4.30475 \dots

Solution steps

\int_{0}^{\frac{3}{2}} 9 - x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{6}} 9 cos^{2} (u) d u

= \int_{0}^{\frac{π}{6}} 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}^{\frac{π}{6}} cos^{2} (u) d u

Rewrite using trig identities

= 9 \cdot \int_{0}^{\frac{π}{6}} \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}^{\frac{π}{6}} 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}^{\frac{π}{6}} 1 d u + \int_{0}^{\frac{π}{6}} cos (2 u) d u)

\int_{0}^{\frac{π}{6}} 1 d u = \frac{π}{6}

\int_{0}^{\frac{π}{6}} cos (2 u) d u = \frac{3}{4}

= 9 \cdot \frac{1}{2} (\frac{π}{6} + \frac{3}{4})

Simplify

9 \cdot \frac{1}{2} (\frac{π}{6} + \frac{3}{4}) : \frac{3 ( 2 π + 3 3 )}{8}

= \frac{3 ( 2 π + 3 3 )}{8}

\int x^{2} (3 x^{2} + 9)^{2} d x

\int csc^{3} (x) cot^{3} (x) d x

x \to \infty lim (\frac{1}{1 0 ^{2^{x}}})

\frac{d}{d j} (- 4 i + 5 j)

cos (x), x = 0

What is the integral from 0 to 3/2 of sqrt(9-x^2) ?
The integral from 0 to 3/2 of sqrt(9-x^2) is (3(2pi+3sqrt(3)))/8