Skills Review for The Definite Integral

Learning Outcomes

Identify the equation of a semicircle
Define a piecewise function or graph from an absolute value function or graph

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 $(x-h)^2+(y-k)^2=r^2$ where $(h,k)$ is the center and $r$ is the radius.

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

$y=\pm \sqrt{r^2-(x-h)^2}+k$

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

Example: Identifying the Equation of a SemiCircle

Identify the center and radius of the semicircle and graph. $y=\sqrt{9-x^2}$

Show Solution

We have been given the equation of a semicircle. Since the radical is positive, this is the upper half of a circle. The radius squared is 9, so this means the radius is 3. The center is $(0,0)$ .

So, the graph of the semicircle is:

The upper half of a circle with center (0,0) and radius 3.

Example: Identifying the Equation of a SemiCircle

Identify the center and radius of the semicircle and graph. $y=-\sqrt{16-(x-2)^2}+1$

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We have been given the equation of a semicircle. Since the radical is negative, this is the lower half of a circle. The radius squared is 16, so this means the radius is 4. The center is $(2,1)$ .

So, the graph of the semicircle is:

The lower half of a circle with center (2,1) and radius 4.

Try It

Identify the center and radius of the semicircle and graph. $y=\sqrt{4-x^2}$

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Define the Absolute Value Function

A General Note: Absolute Value Function

The absolute value function can be defined as a piecewise function

$\lvert x \rvert= \begin{cases} x ,\ x \geq 0 \\ -x , x < 0 \end{cases}$

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

Now, when a variable is involved in the absolute value expression such as $\lvert x-3 \rvert$ , 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: $\lvert x-3 \rvert= \begin{cases} x-3 ,\ x \geq 3 \\ -(x-3) , x < 3 \end{cases}$

Example: Defining the Absolute Value Function

Give the piecewise breakdown of the absolute function $\lvert x-4 \rvert$ .

Show Solution

Notice the absolute value goes from being positive to negative at an x-value of 4.

So, we the piecewise breakdown is:

$\lvert x-4 \rvert= \begin{cases} x-4 ,\ x \geq 4 \\ -(x-4) , x < 4 \end{cases}$

Try It

Give the piecewise breakdown of the absolute function $\lvert 2x-3 \rvert$ .

Show Solution

The piecewise breakdown is: $\lvert 2x-3 \rvert= \begin{cases} 2x-3 ,\ x \geq \frac{3}{2} \\ -(2x-3) , x < \frac{3}{2} \end{cases}$

Integration: Prerequisite Material