What is the integral from 0 to 3 of sqrt(6x-x^2) ?

The integral from 0 to 3 of sqrt(6x-x^2) is (9pi)/4

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integral from 0 to 3 of sqrt(6x-x^2)

\int_{0}^{3} 6 x - x^{2} d x

\frac{9 π}{4}

Solution steps

\int_{0}^{3} 6 x - x^{2} d x

Complete the square

6 x - x^{2} : - (x - 3)^{2} + 9

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

Apply u-substitution

: \int_{- 3}^{0} - u^{2} + 9 d u

= \int_{- 3}^{0} - u^{2} + 9 d u

Apply Trigonometric Substitution

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

= \int_{- \frac{π}{2}}^{0} 9 cos^{2} (v) d v

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} (v) d v

Rewrite using trig identities

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

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 v) d v

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 v + \int_{- \frac{π}{2}}^{0} cos (2 v) d v)

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

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

\int_{- \frac{π}{2}}^{\frac{π}{2}} cos (x) - 1 d x

\int_{0}^{2 π} \frac{1}{2}

\int_{- \frac{π}{2}}^{\frac{π}{2}} cos (x) - 4 d x

\int_{1}^{2} (5 - 2 x) d x

\int_{- 2}^{2} (1 - x^{2}) d x

What is the integral from 0 to 3 of sqrt(6x-x^2) ?
The integral from 0 to 3 of sqrt(6x-x^2) is (9pi)/4