Question

evaluate this integral by using a substitu\\(\int_{0}^{3} x\sqrt{3 - x}\\ dx\\)\\(\int_{0}^{3} x\sqrt{3 - x}\\ dx = \square\\) (type an exact answe

Explanation:

Step1: Set substitution variable

Let $u = 3 - x$, so $x = 3 - u$, and $du = -dx \implies dx = -du$.

Step2: Adjust integral bounds

When $x=0$, $u=3-0=3$; when $x=3$, $u=3-3=0$.

Step3: Rewrite integral in terms of u

$$\int_{0}^{3} x\sqrt{3-x} dx = \int_{3}^{0} (3-u)\sqrt{u} (-du) = \int_{0}^{3} (3u^{\frac{1}{2}} - u^{\frac{3}{2}}) du$$

Step4: Integrate term by term

Use power rule $\int u^n du = \frac{u^{n+1}}{n+1}$:
$$\int_{0}^{3} 3u^{\frac{1}{2}} du - \int_{0}^{3} u^{\frac{3}{2}} du = 3\cdot\frac{u^{\frac{3}{2}}}{\frac{3}{2}} \bigg|_{0}^{3} - \frac{u^{\frac{5}{2}}}{\frac{5}{2}} \bigg|_{0}^{3}$$
Simplify to:
$$2u^{\frac{3}{2}} \bigg|_{0}^{3} - \frac{2}{5}u^{\frac{5}{2}} \bigg|_{0}^{3}$$

Step5: Evaluate at bounds

Substitute $u=3$ and $u=0$:
$$2(3)^{\frac{3}{2}} - 0 - \frac{2}{5}(3)^{\frac{5}{2}} + 0 = 2\cdot3\sqrt{3} - \frac{2}{5}\cdot9\sqrt{3}$$
$$= 6\sqrt{3} - \frac{18}{5}\sqrt{3}$$

Step6: Combine like terms

$$\frac{30}{5}\sqrt{3} - \frac{18}{5}\sqrt{3} = \frac{12}{5}\sqrt{3}$$

Answer:

$\frac{12\sqrt{3}}{5}$