Question

evaluate this integral by using a substitution prior to
\\(\int_{0}^{2} x\sqrt{2 - x} \\, dx\\)
\\(\int_{0}^{2} x\sqrt{2 - x} \\, dx = \square\\) (type an exact answer.)

Explanation:

Step1: Set substitution variable

Let $u = 2 - x$, so $x = 2 - u$, and $du = -dx \implies dx = -du$.

Step2: Adjust integral bounds

When $x=0$, $u=2-0=2$; when $x=2$, $u=2-2=0$.

Step3: Rewrite integral in terms of u

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

Step4: Integrate term by term

Use power rule $\int u^n du = \frac{u^{n+1}}{n+1}$:

$$ \int_{0}^{2} 2u^{\frac{1}{2}} du - \int_{0}^{2} u^{\frac{3}{2}} du = 2\cdot\frac{u^{\frac{3}{2}}}{\frac{3}{2}} \bigg|_{0}^{2} - \frac{u^{\frac{5}{2}}}{\frac{5}{2}} \bigg|_{0}^{2} $$

Simplify coefficients:

$$ \frac{4}{3}u^{\frac{3}{2}} \bigg|_{0}^{2} - \frac{2}{5}u^{\frac{5}{2}} \bigg|_{0}^{2} $$

Step5: Evaluate at bounds

Substitute $u=2$ and $u=0$ (the $u=0$ terms are 0):

$$ \frac{4}{3}(2)^{\frac{3}{2}} - \frac{2}{5}(2)^{\frac{5}{2}} = \frac{4}{3}\cdot2\sqrt{2} - \frac{2}{5}\cdot4\sqrt{2} = \frac{8\sqrt{2}}{3} - \frac{8\sqrt{2}}{5} $$

Step6: Combine fractions

Find common denominator 15:

$$ \frac{40\sqrt{2} - 24\sqrt{2}}{15} = \frac{16\sqrt{2}}{15} $$

Answer:

$\frac{16\sqrt{2}}{15}$