Problem 67 - Calculus: Early Transcendentals

Question

Let \( \displaystyle F(x) = \int^x_2 e^{t^2} \, dt \). Find an equation of the tangent line to the curve \( y = F(x) \) at the point with \( x \)-coordinate 2.

Step-by-Step Solution

Answer

The equation of the tangent line is \( y = e^4 (x - 2) \).

Step 1: Understand the Problem

We are given a function defined by an integral: \( F(x) = \int_{2}^{x} e^{t^2} \, dt \). We need to find the equation of the tangent line to this function at the point where \( x = 2 \).

Step 2: Calculate the Value of the Function at the Point

The function value at the point \( x = 2 \) is calculated by evaluating the integral from 2 to 2. Since the bounds are the same, the integral evaluates to zero: \( F(2) = \int_{2}^{2} e^{t^2} \, dt = 0 \). Thus, the point on the curve is \( (2, 0) \).

Step 3: Use Fundamental Theorem of Calculus for Derivative

According to the Fundamental Theorem of Calculus, if \( F(x) = \int_{a}^{x} f(t) \, dt \), then \( F'(x) = f(x) \). Therefore, \( F'(x) = e^{x^2} \).

Step 4: Find the Slope at the Point

Evaluate \( F'(x) \) at \( x = 2 \) to find the slope of the tangent line at this point. Thus, \( F'(2) = e^{2^2} = e^4 \). The slope of the tangent line is \( e^4 \).

Step 5: Write the Equation of the Tangent Line

The equation of a line with slope \( m \) through a point \( (x_0, y_0) \) is given by \( y - y_0 = m(x - x_0) \). Here, \( m = e^4 \), \( x_0 = 2 \), and \( y_0 = 0 \). Therefore, the equation is \( y - 0 = e^4 (x - 2) \), or simplified, \( y = e^4 (x - 2) \).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus plays a key role in understanding the relationship between differentiation and integration. It connects these two major concepts into a single framework. The theorem states two main things:

• The first part of the theorem tells us that if a function is continuous over an interval, then its integral can be antiderivatized into the original function.
• The second part provides a way to calculate the derivative of an integral function. It states that if you have a function defined by an integral, such as \( F(x) = \int_{a}^{x} f(t) \, dt \), then its derivative is the integrand evaluated at \( x \), i.e., \( F'(x) = f(x) \).

In the given problem, we apply this theorem to derive the function's derivative. By knowing this, we can find out how the function is behaving at any particular point, which is crucial for determining the slope of a tangent line.

Definite Integral

A definite integral is used to calculate the accumulation of quantities, such as areas under a curve, between two specified limits. The integral \( \int_{2}^{x} e^{t^2} \, dt \) is a definite integral where \( t \) is integrated from 2 to \( x \).
When the upper and lower limits of the integral are the same, as they are when \( x = 2 \), the integral evaluates to zero. This is because there is no area to sum between two identical points, which provides a function value of zero at \( x = 2 \).
Thus, knowing how to evaluate a definite integral allows us to pinpoint exact values of a function at specific points for both understanding function behavior and solving problems like finding a tangent line.

Derivative Calculation

The derivative of a function represents its rate of change. By finding the derivative, you can determine the slope of a function at any given point.
In the context of our exercise, differentiating \( F(x) = \int_{2}^{x} e^{t^2} \, dt \) using the Fundamental Theorem of Calculus gives us \( F'(x) = e^{x^2} \). This result tells us how steep the graph is at any specific \( x \) value.
Calculating \( F'(2) \) provides the slope of the tangent line at \( x = 2 \), which is \( e^4 \). This slope is integral to forming the equation of the tangent line, which is next on our agenda.

Slope of a Tangent Line

The concept of a tangent line is fundamental in calculus. It represents a straight line that touches a curve at one point and has the same slope as the curve at that exact point.
To form the equation of this tangent line, we need two pieces of information:

• The slope of the curve at the point (which we've calculated to be \( e^4 \) at \( x = 2 \))
• The actual point of tangency, known here as \( (2, 0) \)

With these, we use the point-slope form of a line equation: \( y - y_0 = m(x - x_0) \). Substituting, we get: \( y - 0 = e^4(x - 2) \).
Thus, the equation of the tangent line is \( y = e^4(x - 2) \), showing how the line aligns with the curve precisely at this point.