Chapter 5: Problem 67
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.
Short Answer
Expert verified
The equation of the tangent line is \( y = e^4 (x - 2) \).
Step by step solution
01
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 \( x = 2 \).
02
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 \( (2, 0) \).
03
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, \( F'(x) = e^{x^2} \).
04
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 \( e^4 \).
05
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, \( y = e^4 (x - 2) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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., \( F'(x) = f(x) \).
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.
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.
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:
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.
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 \( (2, 0) \)
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.