Problem 59 - Essential Calculus

Question

In Exercises 59 and 60 , set up and evaluate the definite integral that gives the area of the region bounded by the graph of the function and the tangent line to the graph at the given point. $$ f(x)=\frac{1}{x^{2}+1}, \quad\left(1, \frac{1}{2}\right) $$

Step-by-Step Solution

Answer

\[ \frac{4- \pi}{4} \]

Step 1: Find the Equation of the Tangent Line

Find the derivative of the given function \(f'(x) = \frac{-2x}{(x^2 + 1)^2}\). At the point (1, 1/2), the derivative \(f'(1) = \frac{-2*1}{(1^2 + 1)^2} = -\frac{1}{2}\), which is the slope of the tangent line. The equation of the line is \(y - y_1= m(x - x_1)\), substituting the values we get equation of tangent line \(y = -\frac{1}{2}(x - 1) + \frac{1}{2}\)

Step 2: Find Intersection Points Between the Function and the Tangent Line

Set the function equal to the tangent line to find the points of intersection: \(\frac{1}{x^2 + 1} = -\frac{1}{2}(x - 1) + \frac{1}{2}\). After a series of simplifications, you can find that the intersection points are x = -1 and x = 1.

Step 3: Set Up and Evaluate the Definite Integral

The definite integral representing the area between the two curves is given by the absolute value of the difference between the function and the tangent line evaluated from -1 to 1: \( \int_{-1}^{1}|\frac{1}{x^2 + 1} + \frac{1}{2}(x - 1) - \frac{1}{2}|dx\). After computation, the value of the integral is \( \frac{4- \pi}{4}\).

Key Concepts

Understanding Tangent Lines

A tangent line is essentially a straight line that just "touches" a curve at a particular point without crossing over it. It is a line that has the same slope as the curve at that point.
In our exercise, the function is given as \( f(x) = \frac{1}{x^2 + 1} \) and the tangent point is \((1, \frac{1}{2})\). To find the tangent line, we first need the derivative of the function. This derivative tells us the slope of the tangent line, which, for complex functions, can be quite valuable in understanding how the function is behaving at specific points.

• The derivative, \( f'(x) = \frac{-2x}{(x^2 + 1)^2} \), gives the slope of the tangent line.
• At the point \((1, \frac{1}{2})\), the slope \( f'(1) = -\frac{1}{2} \).
• Using the point-slope form of the line, the equation of the tangent line becomes \( y = -\frac{1}{2}(x - 1) + \frac{1}{2} \).

This tangent line provides a linear approximation of the curve around \(x = 1\). Understanding it provides a crucial step in analyzing the area between curves.

Calculating Area Between Curves

When calculating the area between two curves, we essentially want to know how much "space" lies between them over a certain interval. This task utilizes definite integrals, which sum up infinitely small areas between the two functions.
In this exercise, we are tasked with finding the area between the curve \( f(x) = \frac{1}{x^2 + 1} \) and its tangent line over the interval from \(x = -1\) to \(x = 1\).

• First, we determine the difference between the function and its tangent line.
• The integral setup is \( \int_{-1}^{1}\left|\frac{1}{x^2 + 1} + \frac{1}{2}(x - 1) - \frac{1}{2}\right|dx \).
• The absolute value is necessary to ensure that the enclosed area is positive, regardless of which curve is on top.

Solving this integral gives the exact area between the function and the tangent line, which is \( \frac{4- \pi}{4} \). This step reveals how integration serves as a powerful tool for capturing portions of space in mathematics.

Locating Intersection Points

Intersection points of functions play a crucial role in many mathematical problems, including area calculations and curve analysis. These are the points where the two graphs touch or cross each other.
For our function \( f(x) = \frac{1}{x^2 + 1} \) and its tangent line, we set them equal to find the intersection points.

• Set the function equal to the tangent line: \( \frac{1}{x^2 + 1} = -\frac{1}{2}(x - 1) + \frac{1}{2} \).
• After algebraic manipulations and simplifications, you derive the solutions for \( x \), which are the intersection points: \( x = -1 \) and \( x = 1 \).

Finding these points confirms the interval over which our area calculations occur, aligning perfectly with the problem's constraints. Intersection points help in determining boundaries for definite integrals and provide understanding where two different curves or behaviors meet.