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

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

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

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

\frac{9 π}{4}

Decimal

7.06858 \dots

Solution steps

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

Apply Trigonometric Substitution

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

Rewrite using trig identities

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

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

\int_{0}^{\frac{π}{2}} 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{d}{d a} (\frac{tan ^{2} ( a ) - sin ^{2} ( a )}{tan ^{2} ( a )})

\frac{d}{d x} (8 x y - \frac{1}{4} (x + y)^{4})

f (x) = x^{2} cos (x^{2})

\int \frac{ln ( x ^{3} )}{x} d x

x \to 2 lim (\frac{2 - ∣ x ∣}{2 + x})

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