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{\partial}{\partial x} (7 y^{6} x^{6} - 6 y^{5} x^{5})

y d x + (x - 4) d y = 0

\frac{d}{d x} (e^{x^{- 2}})

\frac{\partial}{\partial x} (- \frac{4 x}{( x ^{2} - 1 ) ^{2}})

\frac{4}{( s ^{2} + 4 )}

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