Problem 37 - Single Variable Calculus

Question

In Exercises \(9-40\), sketch the region bounded by the graphs of the given equations and find the area of that region. $$ y=\sec ^{2} x, \quad y=2, \quad x=-\frac{\pi}{4}, \quad x=\frac{\pi}{4} $$

Step-by-Step Solution

Answer

The area of the region bounded by the graphs of the given equations \(y = \sec^2x\), \(y = 2\), \(x = -\frac{\pi}{4}\), and \(x = \frac{\pi}{4}\) is \(\frac{2}{3}\) square units.

Step 1: Identify the intersection points

To find the intersection points, we will set $$\sec^2x=2$$ Taking the square root of both sides, we get $$\sec x = \pm\sqrt{2}$$ which is $$\cos x = \frac{1}{\pm\sqrt{2}}$$ So, the intersection points are: $$\cos x = \frac{1}{\sqrt{2}} \Rightarrow x = \pm \frac{\pi}{4}$$

Step 2: Sketch the region

We can now sketch the region bounded by the graphs: 1. Draw the graph of \(y = \sec^2x\). This is a parabolic curve opening upwards. 2. Draw the horizontal line \(y = 2\). 3. Draw the vertical lines \(x = -\frac{\pi}{4}\) and \(x = \frac{\pi}{4}\). 4. Shade the region enclosed by these lines and curves.

Step 3: Compute the area

To compute the area, we'll find the difference between the line and the curve, and then integrate this difference along the interval \(-\frac{\pi}{4} \leq x \leq \frac{\pi}{4} \): $$A = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} [2 - \sec^2x] dx$$ Knowing that \(\sec^2x = 1 + \tan^2x\) and \(\frac{d \tan x}{dx} = \sec^2x\), We get the integral: $$A = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} [1+\tan^2x - \sec^2x] dx$$ $$A = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^2x dx$$ Let \(u = \tan x\), so \(\frac{du}{dx} = \sec^2x\), and the endpoints will be the values of \(u\) for \(x = \pm \frac{\pi}{4}\), thus: $$u = \tan(\frac{\pi}{4}) = 1$$ $$u = \tan(-\frac{\pi}{4}) = -1$$ Now we have: $$A = \int_{-1}^1 u^2 du$$ Calculating the integral gives: $$A = \frac{1}{3}u^3 \Biggr|_{-1}^{1} = \frac{1}{3}(1^3 - (-1)^3) = \frac{2}{3}$$ Therefore, the area of the region is \(\frac{2}{3}\) square units.

Key Concepts

Definite Integration

Definite integration is a fundamental concept in calculus that allows us to calculate the exact area under a curve between two points on a graph. It is performed using the definite integral, which is denoted by an integral sign with upper and lower limits. These limits represent the bounds between which we are finding the area.

For instance, if we want to compute the area under a curve from point 'a' to point 'b', we use the formula \[ A = \int_{a}^{b} f(x)dx \]
where \( f(x) \) is the function representing the curve. The result is a numerical value that gives the net area, taking into account any part of the function that goes below the x-axis, which would reduce the total area. In the case of our exercise, the area in question lies between \( -\frac{\pi}{4} \) and \( \frac{\pi}{4} \). Understanding how to set up and evaluate a definite integral is crucial for finding areas bounded by curves.

Sketching Regions in Calculus

Sketching the region bounded by curves in calculus is a visual approach that aids in understanding the relationship between functions and the area we're interested in. Creating a clear sketch involves several steps:

• Plot each function on a coordinate plane.
• Identify the points where the functions intersect.
• Distinguish the area of interest that is enclosed by the curves and bounds.

These visual cues help in determining how to set up the integral for finding the area. For this exercise, the region is enclosed by the function \( y = \sec^2 x \), the horizontal line \( y = 2 \), and the vertical lines \( x = -\frac{\pi}{4} \) and \( x = \frac{\pi}{4} \). Accurately sketching these on the same axes and shading the bounded region guides us through the integration process. A thorough sketch can also help in identifying symmetry, which can simplify the computation.

Trigonometric Functions Integration

Integration of trigonometric functions is another key area in calculus. Trigonometric functions, such as sine, cosine, and tangent, often appear in calculus problems related to periodic phenomena or circular motion. To integrate these functions, we use their antiderivatives or trigonometric identities to transform the integral into a solvable form.

For example, the identity \( \sec^2 x = 1 + \tan^2 x \) allows us to use the antiderivative of \( \tan x \), which is \( \ln |\sec x| \), or in cases involving \( \tan^2 x \), we might use a substitution as we did in the exercise to simplify the integration. Knowing trigonometric identities is essential in this respect, as they often make the difference between a straightforward problem and one that is not immediately solvable.

Intersection Points of Functions

Finding the intersection points of functions involves solving for the values of 'x' where the two functions have the same 'y' value. Intersection points are crucial for determining the limits of integration when we are computing the area between curves.

When identifying these points, we set the equations of the functions equal to each other and solve for 'x'. Once we have the intersection points, they can serve as the bounds for our definite integral. In our exercise with the functions \( y = \sec^2 x \) and \( y = 2 \), the intersection occurred when \( \sec^2 x = 2 \), leading to the solutions for 'x' at \( \pm \frac{\pi}{4} \). These intersection points act as the limits for integration in our problem. Without accurately determining these points, we might not compute the correct area.