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

The integral from-3 to 0 of sqrt(9-x^2) is (9pi)/4

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

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

\frac{9 π}{4}

Decimal Notation

7.06858 \dots

Solution steps

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

\int_{- \frac{π}{2}}^{0} 1 d u = \frac{π}{2}

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

= 9 \cdot \frac{1}{2} (\frac{π}{2} + 0)

Simplify

9 \cdot \frac{1}{2} (\frac{π}{2} + 0) : \frac{9 π}{4}

= \frac{9 π}{4}

\frac{3 s - 1}{( s ^{2} + 2 s + 5 )}

\frac{d y}{d x} = 3 (y^{2} + 1), y (\frac{π}{4}) = 1

y^{^{'}} - y = 2 t e^{2 t}

\frac{\partial}{\partial y} (2 x e^{x^{2} - y^{2}})

\int y e^{- y} d y

What is the integral from-3 to 0 of sqrt(9-x^2) ?
The integral from-3 to 0 of sqrt(9-x^2) is (9pi)/4