QuestionEvaluate the integral $\int_{0}^{a} x \sqrt{a^{2}-x^{2}} \, dx$ .

Studdy Solution

STEP 1

Assumptions1. The integral is definite from0 to a. . The function to be integrated is $x \sqrt{a^{}-x^{}}$ .
3. We are integrating with respect to $x$ .

STEP 2

The function is a product of $x$ and $\sqrt{a^{2}-x^{2}}$ . This suggests that we should use the method of integration by substitution. We will let $u = a^{2}-x^{2}$ .
$u = a^{2}-x^{2}$

STEP 3

Differentiate $u$ with respect to $x$ to get $du/dx$ .
$\frac{du}{dx} = -2x$

STEP 4

Rearrange the above equation to express $dx$ in terms of $du$ .
$dx = -\frac{du}{2x}$

STEP 5

Substitute $u$ and $dx$ into the integral.
$\int_{0}^{a} x \sqrt{u} (-\frac{du}{2x})$

STEP 6

implify the integral. The $x$ in the numerator and denominator cancel out.
$-\frac{1}{2} \int_{0}^{a} \sqrt{u} du$

STEP 7

The limits of the integral are still in terms of $x$ . We need to change them to be in terms of $u$ . When $x=0$ , $u=a^{2}$ and when $x=a$ , $u=0$ . So the limits of the integral become from $a^{2}$ to $0$ .
$-\frac{1}{2} \int_{a^{2}}^{0} \sqrt{u} du$

STEP 8

The limits of the integral are in the wrong order. We can switch them and remove the negative sign.
$\frac{1}{2} \int_{0}^{a^{2}} \sqrt{u} du$

STEP 9

Now, we can integrate $\sqrt{u}$ with respect to $u$ .
$\frac{}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{}^{a^{2}}$

STEP 10

Substitute the limits of integration into the antiderivative.
$\frac{}{2} [\frac{2}{3} (a^{2})^{\frac{3}{2}} - \frac{2}{3} (0)^{\frac{3}{2}}]$

STEP 11

implify the expression.
$\frac{}{} [\frac{}{3} a^{3} -0]$

STEP 12

implify further to get the final answer.
$\frac{}{} a^{}$ So, the solution to the integral $\int_{0}^{a} x \sqrt{a^{2}-x^{2}} d x$ is $\frac{}{} a^{}$ .

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