Chapter 6: Problem 57
Use a graphing calculator or a CAS to do. Find the area of the region between the graphs of \(y=f(x)=\exp \left(-x^{2}\right)\) and \(y=f^{\prime \prime}(x)\) on [-3,3].
Short Answer
Expert verified
The area between the two graphs from \([-3, 3]\) is approximately using a calculator or a CAS.
Step by step solution
01
Understand the functions involved
We start with the function \( y = f(x) = e^{-x^2} \). We also need to find its second derivative, \( f''(x) \), for the comparison.
02
Compute the first derivative
The first derivative of \( f(x) = e^{-x^2} \) is found using the chain rule. Let \( u = -x^2 \), then \( f'(x) = \frac{d}{dx}[e^u] \cdot \frac{du}{dx} = e^{-x^2} \cdot (-2x) = -2x e^{-x^2} \).
03
Compute the second derivative
Differentiate \( f'(x) = -2x e^{-x^2} \) again using the product rule: \( f''(x) = \frac{d}{dx}[-2x e^{-x^2}] = -2(e^{-x^2} + 2x^2 e^{-x^2}) = -2 e^{-x^2} (1 - 2x^2) \).
04
Set up the integral to find the area
To find the area between \( f(x) \) and \( f''(x) \), integrate the absolute difference between these functions over \([-3, 3]\): \( \int_{-3}^{3} |e^{-x^2} - (-2 e^{-x^2} (1 - 2x^2))| \, dx \).
05
Simplify and calculate the integral
Simplify the integrand: \( |e^{-x^2} + 2e^{-x^2} (1 - 2x^2)| = |e^{-x^2} (1 + 2 - 4x^2)| = |e^{-x^2} (3 - 4x^2)| \). Use a graphing calculator or a computer algebra system (CAS) to compute the integral \( \int_{-3}^{3} e^{-x^2} |3 - 4x^2| \, dx \), which gives the total area.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
The concept of a definite integral is fundamental when calculating the area between curves. The definite integral is represented as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration. This integral calculates the net area between the function \( f(x) \) and the x-axis over the interval \([a, b]\).
In the context of finding the area between two curves, like \( y = f(x) \) and \( y = g(x) \), you calculate the definite integral of their absolute difference: \( \int_{a}^{b} |f(x) - g(x)| \, dx \). This accounts for any regions where one function dips below the other.
In the context of finding the area between two curves, like \( y = f(x) \) and \( y = g(x) \), you calculate the definite integral of their absolute difference: \( \int_{a}^{b} |f(x) - g(x)| \, dx \). This accounts for any regions where one function dips below the other.
- Ensure correct setup of the integral limits according to the intersection points of functions.
- Simplify the integrand if possible to make calculation straightforward, especially if using a calculator.
Second Derivative
The second derivative of a function provides significant insights, particularly in motion and curve sketching. It is obtained by differentiating the first derivative \( f'(x) \) again.
In the example given, the function is \( y = e^{-x^2} \), whose second derivative was calculated as \( f''(x) = -2 e^{-x^2} (1 - 2x^2) \).
The second derivative is useful for determining:
In the example given, the function is \( y = e^{-x^2} \), whose second derivative was calculated as \( f''(x) = -2 e^{-x^2} (1 - 2x^2) \).
The second derivative is useful for determining:
- Concavity of the function: Whether the function is curving upwards or downwards on a particular interval.
- Points of inflection: Where the concavity changes, indicating a potential change in the behavior of the function curve.
Graphing Calculator
A graphing calculator or computer algebra system (CAS) plays an essential role in evaluating complex integrals and visualizing functions. These tools help simplify calculations and provide a visual representation of the functions involved, making it easier to understand their behaviors.
For instance, when calculating the area between the curves \( y = f(x) \) and \( y = f''(x) \) over the interval \([-3, 3]\), plugging the simplified integrand into a graphing calculator ensures efficient computation. Especially if the integrand is complex, like in \( \int_{-3}^{3} e^{-x^2} |3 - 4x^2| \, dx \), manually solving might be impractical.
Key benefits of using a graphing calculator include:
For instance, when calculating the area between the curves \( y = f(x) \) and \( y = f''(x) \) over the interval \([-3, 3]\), plugging the simplified integrand into a graphing calculator ensures efficient computation. Especially if the integrand is complex, like in \( \int_{-3}^{3} e^{-x^2} |3 - 4x^2| \, dx \), manually solving might be impractical.
Key benefits of using a graphing calculator include:
- Rapidly computing integral values that are otherwise difficult to evaluate analytically.
- Offering graphical insights, which can help in confirming the relationship and positions of functions.
Chain Rule
The Chain Rule is a crucial differentiation technique, particularly when dealing with composite functions. It helps find the derivative of a function that is composed of two or more functions.
In our example, we needed to find the first derivative of \( y = e^{-x^2} \). The Chain Rule comes into play as follows: Letting \( u = -x^2 \), the derivative becomes \( \frac{d}{dx}[e^u] \cdot \frac{du}{dx} \), simplifying to \( -2x e^{-x^2} \).
The Chain Rule allows us to:
In our example, we needed to find the first derivative of \( y = e^{-x^2} \). The Chain Rule comes into play as follows: Letting \( u = -x^2 \), the derivative becomes \( \frac{d}{dx}[e^u] \cdot \frac{du}{dx} \), simplifying to \( -2x e^{-x^2} \).
The Chain Rule allows us to:
- Differentiate effectively when dealing with nested functions.
- Break down complex differentiation problems into simpler, manageable parts.
Product Rule
When you need to differentiate a function that is the product of two other functions, the Product Rule is your tool of choice. It simplifies the process by providing a formula to differentiate products of functions.
The formula for the Product Rule is \( (fg)' = f'g + fg' \), where \( f \) and \( g \) are functions of x. In our specific case, to get the second derivative \( f''(x) \), we apply the Product Rule to \( f'(x) = -2x e^{-x^2} \):
- Differentiate \( f(x) = -2x \) which is \( -2 \).
- Differentiate \( g(x) = e^{-x^2} \) using the chain rule which results in \( -2x e^{-x^2} \).
Then apply \( f''(x) = (-2)e^{-x^2} + (-2x)(-2x e^{-x^2}) \), simplifying to \( -2 e^{-x^2} (1 - 2x^2) \).
The Product Rule is efficient for:
The formula for the Product Rule is \( (fg)' = f'g + fg' \), where \( f \) and \( g \) are functions of x. In our specific case, to get the second derivative \( f''(x) \), we apply the Product Rule to \( f'(x) = -2x e^{-x^2} \):
- Differentiate \( f(x) = -2x \) which is \( -2 \).
- Differentiate \( g(x) = e^{-x^2} \) using the chain rule which results in \( -2x e^{-x^2} \).
Then apply \( f''(x) = (-2)e^{-x^2} + (-2x)(-2x e^{-x^2}) \), simplifying to \( -2 e^{-x^2} (1 - 2x^2) \).
The Product Rule is efficient for:
- Differentiating functions that are non-linearly intertwined.
- Complex expressions where manual differentiation could lead to mistakes.