Problem 12 - Calculus Single Variable

Question

A function \(f\) is defined on a specified interval \(I=[a, b] .\) Calculate the area of the region that lies between the vertical lines \(x=a\) and \(x=b\) and between the graph of \(f\) and the \(x\) -axis. $$ f(x)=8 x-5 /\left(1+x^{2}\right) \quad I=[-1,2] $$

Step-by-Step Solution

Answer

The area under the curve from \( x = -1 \) to \( x = 2 \) is approximately 10.39.

Step 1: Understand the Problem

We need to calculate the area under the curve of the function \( f(x) = 8x - \frac{5}{1+x^2} \) from \( x = -1 \) to \( x = 2 \). This means we are looking for the definite integral of \( f(x) \) over the interval \([-1, 2]\).

Step 2: Set Up the Integral

The area under the curve can be found by calculating the integral: \[\int_{-1}^{2} \left(8x - \frac{5}{1+x^2}\right) \, dx\]This represents the area between the graph of \( f \) and the x-axis from \( x = -1 \) to \( x = 2 \).

Step 3: Integrate Term-by-Term

The integral can be separated and solved as two separate integrals: \[\int_{-1}^{2} 8x \, dx - \int_{-1}^{2} \frac{5}{1+x^2} \, dx\]Compute each integral separately.

Step 4: Integrate 8x

To integrate \( 8x \), use the power rule: \[\int 8x \, dx = 4x^2 + C\]Thus, \[\int_{-1}^{2} 8x \, dx = \left[4x^2\right]_{-1}^{2} = 4(2)^2 - 4(-1)^2 = 16 - 4 = 12\]

Step 5: Integrate \(-\frac{5}{1+x^2}\)

To integrate \( -\frac{5}{1+x^2} \), recognize it as a standard derivative:\[\int -\frac{5}{1+x^2} \, dx = -5 \arctan(x) + C\]Thus, \[\int_{-1}^{2} -\frac{5}{1+x^2} \, dx = \left[-5 \arctan(x)\right]_{-1}^{2} = -5 \arctan(2) + 5 \arctan(-1)\]

Step 6: Evaluate the ArcTan Values

Compute the values of \( \arctan(2) \) and \( \arctan(-1) \):- \( \arctan(2) \) is the angle whose tangent is 2. Common approximate value = 1.107 (in radians).- \( \arctan(-1) = -\frac{\pi}{4}\) because tangent of \((-\frac{\pi}{4})\) is -1. Substitute these values back:\[-5(1.107) + 5(-\frac{\pi}{4}) \approx -5.535 + \frac{5\pi}{4}\]

Step 7: Complete the Calculation

Sum the results of the two integrals obtained in steps 4 and 5:\[12 + (-5.535 + \frac{5\pi}{4}) \approx 12 - 5.535 + \frac{5\pi}{4}\]This results in the approximate area under the curve.

Step 8: Final Area Value

Calculate the exact numerical value by substituting \(\pi \approx 3.14159\):\[12 - 5.535 + 3.92699 \approx 10.392\]This gives us the area under the curve from \( x = -1 \) to \( x = 2 \).

Key Concepts

Area Under the Curve

Calculating the area under a curve is a fundamental concept in calculus. It involves finding the core measurement of a region between a given function and the x-axis across a specified interval. For the function given, the interval is from \( x = -1 \) to \( x = 2 \). This area is computed using definite integrals.

The definite integral represents the sum of infinitesimally small quantities, which resemble the area of thin vertical slices under the curve. By summing these vertical slice areas from \( x = -1 \) to \( x = 2 \), we obtain the total area. This is expressed mathematically as:

• The integral of \( f(x) = 8x - \frac{5}{1+x^2}\) over the interval \([-1, 2]\)
• Symbolized as \[ \int_{-1}^{2} \left(8x - \frac{5}{1+x^2}\right) \, dx \]

Each of the terms in this function is integrated separately to simplify the process.

Integration Techniques

Integration techniques involve a variety of strategies to compute integrals effectively. In this exercise, the primary approach is to split a complex integral into simpler parts. Each part can then be integrated separately using standard calculus techniques.

For our function \( f(x) = 8x - \frac{5}{1+x^2} \), we use:

• The power rule for \( 8x \), where \( \int 8x \, dx = 4x^2 + C \)
• Recognizing the term \( \frac{5}{1+x^2} \) as a derivative of an arctangent function

These strategies allow us to tackle each part efficiently:

• First, integrate \( 8x \) over \([-1, 2]\) to obtain \( [4x^2]_{-1}^{2} \)
• Next, integrate \( -\frac{5}{1+x^2} \) as \( -5 \arctan(x) \) to find \( [-5 \arctan(x)]_{-1}^{2} \)

By knowing these standard techniques, one can solve many integration problems by recognizing common integral forms and applying appropriate methods.

Arc Tangent Function

The arc tangent function, denoted as \( \arctan(x) \), is an important trigonometric function often used in calculus when dealing with integrals involving the term \( \frac{1}{1+x^2} \). This arises because the derivative of \( \arctan(x) \) is \( \frac{1}{1+x^2} \).

When solving the integral \( \int -\frac{5}{1+x^2} \, dx \), it becomes \( -5 \arctan(x) + C \). This is because each component within the integral corresponds to a known derivative form, making integration straightforward.

To evaluate at bounds, compute:

• \( \arctan(2) \), which means finding the angle whose tangent is 2. Approximated as 1.107 radians
• \( \arctan(-1) = -\frac{\pi}{4} \), as the tangent of \(-\frac{\pi}{4}\) is -1

By combining these evaluations, we calculate the definite arc tangent contribution to the area under the curve, aiding in reaching the final area value for the region defined by our integral.