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Chapter 4: Problem 32

Evaluating a Definite Integral Using a Geometric Formula In Exercises \(23-32\) , sketch the region whose area is given by the definite integral. Then use a geometric formula to evaluate the integral \((a>0, r>0).\) $$\int_{-r}^{r} \sqrt{r^{2}-x^{2}} d x$$

Short Answer

Expert verified
The value of the integral \(\int_{-r}^{r} \sqrt{r^{2}-x^{2}} d x\) is \( \frac{1}{2}\pi r^{2} \).

Step by step solution

01

Sketch of the region

The region represented by the integral is the area enclosed by the curve of the function \( y=\sqrt{r^{2}-x^{2}} \) and the x-axis from \( x=-r \) to \( x= r \). This function and boundary form a semicircle of radius \( r \) above the x-axis.
02

Application of geometric formula

According to standard formulas, the area \( A \) of a circle with radius \( r \) is given by \( A=\pi r^{2} \). However, the area calculated by the integral is a semicircle's area, which is half of the area of a circle. Thus, the area of the region is \( A=\frac{1}{2}\pi r^{2} \).
03

Evaluation of the integral

According to the sketch and geometric analysis, the value of the integral \(\int_{-r}^{r} \sqrt{r^{2}-x^{2}} d x\) is the area \( \frac{1}{2}\pi r^{2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Geometric Formula
In the realm of calculus, geometric formulas often simplify the process of evaluating integrals. When dealing with definite integrals representing familiar shapes, you can use geometric formulas to find area more efficiently.
For circles and semicircles, well-known geometric formulas are invaluable for quick calculations.
  • The area of a full circle: \( \pi r^2 \)
  • The area of a semicircle (half of a circle): \( \frac{1}{2}\pi r^2 \)
Leveraging these relationships, you can substitute numerical integration with a straightforward arithmetic approach, provided the integral describes a standard geometric shape.
Semicircle Area
When you encounter an integral that results in a semicircular region, understanding the area calculation is key. The function \( y = \sqrt{r^2 - x^2} \) represents the upper half of a circle centered on the origin.
It spans from \( x = -r \) to \( x = r \), creating a semicircle above the x-axis.
To find this area using a geometric approach:
  • Recognize the entire circle's area: \( \pi r^2 \)
  • Calculate half, suitable for the semicircle: \( \frac{1}{2} \pi r^2 \)
This method simplifies the task, providing you with a neat solution without extensive calculations.
Integral Evaluation
Integral evaluation through geometric perspective is powerful when the function and limits suggest a common shape.
For the integral \( \int_{-r}^{r} \sqrt{r^2 - x^2} \, dx \), sketching reveals it covers a semicircle above the x-axis.
Instead of calculating mathematically using limits and function behavior, you directly calculate the area using the semicircle's properties.
This leads to an area of \( \frac{1}{2} \pi r^2 \), demonstrating the efficiency of recognizing geometric figures in definite integrals.
  • Sketch to identify shapes
  • Apply geometric formulas
  • Relate back to definite integral value
Recognizing such patterns helps in tackling similar integrals without unnecessary complexity.

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Most popular questions from this chapter

In Exercises 109 and 110, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. \begin{equation} \begin{array}{l}{\text { If } F(b)-F(a)=G(b)-G(a), \text { then } F^{\prime}(x)=G^{\prime}(x) \text { on the }} \\ {\text { interval }[a, b] .}\end{array} \end{equation}

Using the Second Fundamental Theorem of Calculus In Exercises 75-80, use the Second Fundamental Theorem of Calculus to find \(F^{\prime}(x).\) $$F(x)=\int_{-1}^{x} \sqrt{t^{4}+1} d t$$

Using an Even Function Use \(\int_{0}^{6} x^{2} d x=72\) to evaluate each definite integral without using the Fundamental Theorem of Calculus. \(\begin{array}{ll}{\text { (a) } \int_{-6}^{6} x^{2} d x} & {\text { (b) } \int_{-6}^{0} x^{2} d x} \\ {\text { (c) } \int_{0}^{6}-2 x^{2} d x} & {\text { (d) } \int_{-6}^{6} 3 x^{2} d x}\end{array}\)

Even and Odd Functions In Exercises 79 and 80, write the integral as the sum of the integral of an odd function and the integral of an even function. Use this simplification to evaluate the integral. $$\int_{-3}^{3}\left(x^{3}+4 x^{2}-3 x-6\right) d x$$

Modeling Data An experimental vehicle is tested on a straight track. It starts from rest, and its velocity v (in meters per second) is recorded every 10 seconds for 1 minute (see table). $$\begin{array}{|c|c|c|c|c|c|c|}\hline t & {0} & {10} & {20} & {30} & {40} & {50} & {60} \\ \hline v & {0} & {5} & {21} & {40} & {62} & {78} & {83} \\\ \hline\end{array}$$ \begin{equation} \begin{array}{l}{\text { (a) Use a graphing utility to find a model of the form }} \\ {v=a t^{3}+b t^{2}+c t+d \text { for the data. }} \\ {\text { (b) Use a graphing utility to plot the data and graph the model. }} \\ {\text { (c) Approximate the distance traveled by the vehicle during }} \\ {\text { the test. }}\end{array} \end{equation}
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