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Find the areas of the regions enclosed by the lines and curves in Exercises \(63-70 .\) $$ y=\sec ^{2} x, \quad y=\tan ^{2} x, \quad x=-\pi / 4, \quad \text { and } \quad x=\pi / 4 $$

Short Answer

Expert verified
The area is \( \frac{\pi}{2} \).

Step by step solution

01

Set Up the Problem

We need to determine the area enclosed by the curves \( y = \sec^2 x \) and \( y = \tan^2 x \) between the vertical lines \( x = -\frac{\pi}{4} \) and \( x = \frac{\pi}{4} \). We will do this by finding the area between the curves across the specified interval.
02

Identify the Points of Intersection

For the functions \( y = \sec^2 x \) and \( y = \tan^2 x \), observe that the curves intersect at the points \( x = -\frac{\pi}{4} \) and \( x = \frac{\pi}{4} \).
03

Determine which Functions to Integrate

Since \( \sec^2 x > \tan^2 x \) for all \( x \) in \([-\frac{\pi}{4}, \frac{\pi}{4}] \), we will integrate \( \sec^2 x - \tan^2 x \) over this interval to find the area between the two curves.
04

Set Up the Integral

The area \( A \) is given by the integral: \[A = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\sec^2 x - \tan^2 x) \, dx\]Simplify inside the integral using the identity \( \sec^2 x - \tan^2 x = 1 \).
05

Evaluate the Integral

Now, evaluate the integral \[\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 1 \, dx\]This becomes \( x \) evaluated from \(-\frac{\pi}{4}\) to \( \frac{\pi}{4} \).
06

Compute the Final Result

Substitute in the limits of integration: \[A = \left[ x \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2}\]Thus, the area enclosed by the curves is \( \frac{\pi}{2} \).

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Key Concepts

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

Area Between Curves
The concept of the area between curves is crucial in integral calculus. When you have two functions, the area between them can be found by integrating the difference of these functions over a specific interval. In our exercise, we need to determine the region enclosed between the two curves, \( y = \sec^2 x \) and \( y = \tan^2 x \), within the vertical boundaries of \( x = -\frac{\pi}{4} \) and \( x = \frac{\pi}{4} \).To find the exact area:
  • First, identify the points where the curves intersect or the given interval over which you will integrate.
  • Next, set up an integral where the upper function minus the lower function is the integrand. In the given problem, \( \sec^2 x \) is always greater than \( \tan^2 x \) in the specified interval.
  • The definite integral \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\sec^2 x - \tan^2 x) \, dx \) gives the area.
Understanding how to calculate areas between curves allows us to visualize and comprehend the spaces that shapes occupy, a core aspect of many calculus applications.
Definite Integrals
Definite integrals are powerful tools in calculus that provide the total accumulation of quantities, often resulting in finding areas under curves. A definite integral is expressed as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the interval limits.When working with definite integrals:
  • The lower limit \( a \) and the upper limit \( b \) define the interval over which the integration occurs. They can represent physical boundaries, such as time or space.
  • In the context of area under a curve, the definite integral of a function can reveal the area enclosed between the curve and the x-axis.
In our problem, simplifying under the integral was possible due to the trigonometric identity \( \sec^2 x - \tan^2 x = 1 \), making the integration straightforward. Evaluating \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 1 \, dx \), we find the enclosed area by simply computing the difference from the upper and lower limits, resulting in \( \frac{\pi}{2} \). This illustrates how definite integrals can simplify problem-solving by breaking down complex expressions into tangible areas.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that hold true for every value within their domains. They are essential for simplifying expressions and solving trigonometric integrals, as seen in this exercise with identities like \( \sec^2 x - \tan^2 x = 1 \).These identities serve several purposes:
  • They simplify complex expressions, making integration more straightforward.
  • Provide relationships among trigonometric functions, useful in solving various calculus problems.
In the context of our problem, understanding the identity \( \sec^2 x = 1 + \tan^2 x \) allowed the simplification of the integral \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\sec^2 x - \tan^2 x) \, dx \) to \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 1 \, dx \). This reduced the task to a simple calculation of integration limits, showcasing how identities can offer elegant shortcuts in calculus solutions.

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Most popular questions from this chapter

Use the Substitution Formula in Theorem 6 to evaluate the integrals in Exercises \(1-24 .\) $$ \int_{0}^{1} \sqrt{t^{5}+2 t}\left(5 t^{4}+2\right) d t $$

In Exercises \(91-94\) , you will find the area between curves in the plane when you cannot find their points of intersection using simple algebra. Use a CAS to perform the following steps: a. Plot the curves together to see what they look like and how many points of intersection they have. b. Use the numerical equation solver in your CAS to find all the points of intersection. c. Integrate \(|f(x)-g(x)|\) over consecutive pairs of intersection values. d. Sum together the integrals found in part (c). $$ f(x)=\frac{x^{4}}{2}-3 x^{3}+10, \quad g(x)=8-12 x $$

Find the areas of the regions enclosed by the lines and curves in Exercises \(51-58 .\) $$ x=y^{3}-y^{2} \quad \text { and } \quad x=2 y $$

The region bounded below by the parabola \(y=x^{2}\) and above by the line \(y=4\) is to be partitioned into two subsections of equal area by cutting across it with the horizontal line \(y=c\) . a. Sketch the region and draw a line \(y=c\) across it that looks about right. In terms of \(c,\) what are the coordinates of the points where the line and parabola intersect? Add them to your figure. b. Find \(c\) by integrating with respect to \(y .\) (This puts \(c\) in the limits of integration.) c. Find \(c\) by integrating with respect to \(x\) . (This puts \(c\) into the integrand as well.)

Solve the initial value problems in Exercises \(53-58\). $$ \frac{d s}{d t}=8 \sin ^{2}\left(t+\frac{\pi}{12}\right), \quad s(0)=8 $$

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