Problem 20

Question

In Problems 17-22, calculate $\int_{a}^{b} f(x) d x$, where a and $b$ are the left and right end points for which $f$ is defined, by using the Interval Additive Property and the appropriate area formulas from plane geometry. Begin by graphing the given function. \(f(x)= \begin{cases}-\sqrt{4-x^{2}} & \text { if }-2 \leq x \leq 0 \\ -2 x-2 & \text { if } 0

Step-by-Step Solution

Verified

Answer

The integral is $-2\pi - 4$.

1Step 1: Graph the Function

Begin by drawing the graph of the piecewise function. We have two parts to this function. The first part, $f(x) = -\sqrt{4-x^2}$ for $-2 \leq x \leq 0$, is the lower half of a circle with a radius of 2 centered at the origin. The second part, $f(x) = -2x - 2$ for $0 < x \leq 2$, is a straight line with a slope of -2 and a y-intercept of -2. Plot these carefully on your coordinate plane.

2Step 2: Recognize Geometric Shapes

Identify the geometric shapes formed by the graph in each interval. From $-2$ to $0$, the graph is a semicircle below the x-axis. From $0$ to $2$, the graph is a right triangle below the x-axis.

3Step 3: Compute Area of Semicircle

Calculate the area of the semicircle. The formula for the area of a full circle is $\pi r^2$. Therefore, the area of the semicircle is $-\frac{1}{2} \times \pi \times 2^2 = -2\pi$. The area is negative because it is below the x-axis.

4Step 4: Compute Area of Triangle

Calculate the area of the triangle. The formula for the area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$. Here, the base is 2 and the height is 4 $(0,-2) \text{ to } (2,-6)$. Thus, the area is $-\frac{1}{2} \times 2 \times 4 = -4$, with the area negative since it's below the x-axis.

5Step 5: Use Interval Additive Property

Compute the integral using the interval additive property. Add the areas of the semicircle and the triangle: $-2\pi + (-4) = -2\pi - 4$.

Key Concepts

Piecewise FunctionsGeometric IntegrationPlane GeometryInterval Additive Property

Piecewise Functions

Piecewise functions are a type of function defined by different expressions depending on the interval of the input value. In our exercise, the function $f(x)$ is defined differently for two different intervals.

For $-2 \leq x \leq 0$, the function is represented by $-\sqrt{4-x^2}$, which describes the lower half of a circle with a radius of 2.
For $0 < x \leq 2$, the function is given by $-2x - 2$, forming a straight line.

The key to working with piecewise functions is to handle each part separately while considering their respective intervals. This way, we can accurately interpret the behavior of the function across its entire domain. Graphing each segment also helps visualize how these parts connect smoothly within the specified intervals.

Geometric Integration

Geometric integration is a technique that involves finding the area under a curve by identifying geometric shapes within the graphs of functions. Instead of relying on purely algebraic methods, we use known area formulas from geometry.
In this exercise, each part of the piecewise function corresponds to a basic geometric shape:

From $-2$ to $0$, the graph of $-\sqrt{4-x^2}$ forms a semicircle.
From $0$ to $2$, the graph of $-2x - 2$ forms a right triangle.

By calculating the area of these shapes and summing them, you can determine the definite integral over the entire interval. This method leverages our understanding of geometry to simplify the calculation process, especially for complex shapes.

Plane Geometry

Plane geometry is the branch of mathematics that deals with shapes and figures on a flat surface, such as lines, triangles, circles, and more. It's essential in geometric integration as it provides the tools and formulas to calculate areas.
To solve our exercise:

For the semicircle, use the area formula for a circle: $\pi r^2$. Since it's a semicircle, the area is $-\frac{1}{2} \times \pi \times 2^2 = -2\pi$.
The triangle's area can be obtained using the formula: $\frac{1}{2} \times \text{base} \times \text{height}$. Here, the base is 2 and the height is 4, giving an area of $-\frac{1}{2} \times 2 \times 4 = -4$.

The negative sign indicates areas below the x-axis in our coordinate plane.

Interval Additive Property

The interval additive property in calculus allows us to split the definite integral of a function over an interval into the sum of integrals over subintervals. This property makes it easier to handle complex functions by breaking them into manageable parts.
In our exercise, the integral from $-2$ to $2$ of $f(x)$ was divided into two parts:

The interval $[-2, 0]$, where the function describes a semicircle.
The interval $[0, 2]$, where the function forms a triangle.

By calculating the area for each interval separately and then adding them together, you apply the interval additive property. This approach simplifies integration, especially for piecewise functions.

Problem 19

Problem 20

Other exercises in this chapter

Problem 19

If a function $f$ is increasing on $[a, b]$, will the left Riemann sum be larger or smaller than $\int_{a}^{b} f(x) d x$ ?

View solution

Problem 19

In Problems 17-26, find $G^{\prime}(x)$. $$ G(x)=\int_{0}^{x}\left(2 t^{2}+\sqrt{t}\right) d t $$

View solution

Problem 20

In Problems 15-34, use the method of substitution to find each of the following indefinite integrals. $\int \cos (\pi v-\sqrt{7}) d v$

View solution

Problem 20

If a function $f$ is increasing on $[a, b]$, will the right Riemann sum be larger or smaller than $\int_{a}^{b} f(x) d x$ ?

View solution