In the Definite Integral section, we will learn how to find the area under a curve using integrals. Here we will review the equation of a semicircle, a curve we will often find the area under. We will also review the mathematical definition of the absolute value function, a skill needed to evaluate definite integrals that contain such a function.

Identify the Equation of a Semicircle

The standard form of a circle is [latex](x-h)^2+(y-k)^2=r^2[/latex] where [latex](h,k)[/latex] is the center and [latex]r[/latex] is the radius.

If we take the standard form equation of a circle and solve for y, we obtain:

[latex]y=\pm \sqrt{r^2-(x-h)^2}+k[/latex]

[latex]y= \sqrt{r^2-(x-h)^2}+k[/latex] is the upper half of the circle and [latex]y=- \sqrt{r^2-(x-h)^2}+k[/latex] is the lower half of the circle.

Define the Absolute Value Function

Therefore, when evaluating an absolute value expression such as [latex]\lvert -9 \rvert[/latex], the true way to evaluate is as follows: [latex]\lvert -9 \rvert= -(-9) = 9[/latex]

Now, when a variable is involved in the absolute value expression such as [latex]\lvert x-3 \rvert[/latex], notice what is inside the absolute value goes from being positive to negative at an x-value of 3. The piecewise breakdown of this absolute value function is as follows: [latex]\lvert x-3 \rvert= \begin{cases} x-3 ,\ x \geq 3 \\ -(x-3) , x < 3 \end{cases}[/latex]