What is the integral from 0 to 3 of xsqrt(3-x) ?

The integral from 0 to 3 of xsqrt(3-x) is -(18sqrt(3))/5+6sqrt(3)

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integral from 0 to 3 of xsqrt(3-x)

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

- \frac{18 3}{5} + 63

Decimal Notation

4.15692 \dots

Solution steps

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

Apply u-substitution

: \int_{3}^{0} - (- u + 3) u d u

= \int_{3}^{0} - (- u + 3) u d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{3} - (- u + 3) u d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{0}^{3} (- u + 3) u d u)

Expand

(- u + 3) u : - u^{\frac{3}{2}} + 3 u

= - (- \int_{0}^{3} - u^{\frac{3}{2}} + 3 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

= - (- (- \int_{0}^{3} u^{\frac{3}{2}} d u + \int_{0}^{3} 3 u d u))

\int_{0}^{3} u^{\frac{3}{2}} d u = \frac{18 3}{5}

\int_{0}^{3} 3 u d u = 63

= - (- (- \frac{18 3}{5} + 63))

Simplify

= - \frac{18 3}{5} + 63

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\int_{- 1}^{1} x^{3} - 2 x^{2} - 5 x + 6 d x

What is the integral from 0 to 3 of xsqrt(3-x) ?
The integral from 0 to 3 of xsqrt(3-x) is -(18sqrt(3))/5+6sqrt(3)