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Chapter 6: Problem 48

Give an example of: A function \(G(x),\) constructed using the Second Fundamental Theorem of Calculus, such that \(G\) is concave up and \(G(7)=0\).

Short Answer

Expert verified
The function is \( G(x) = \frac{1}{2}x^2 - \frac{49}{2} \).

Step by step solution

01

Understand the Second Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus states that if \( f \) is a continuous function on \([a, b]\), and \( G(x) = \int_a^x f(t) \, dt \), then \( G'(x) = f(x) \). To find a function that is concave up, we need \( G''(x) > 0 \).
02

Choose a Suitable Function for Concavity

To ensure that \( G(x) \) is concave up, choose \( f(x) \) such that its derivative, \( f'(x) = G''(x) \), is positive. A simple choice is \( f(x) = x \), since \( G''(x) = f'(x) = 1 > 0 \).
03

Construct the Function Using Integration

Using the chosen function \( f(x) = x \), construct \( G(x) \):\[G(x) = \int_a^x t \, dt = \left[ \frac{1}{2}t^2 \right]_a^x = \frac{1}{2}x^2 - \frac{1}{2}a^2.\]
04

Ensure the Function Satisfies the Condition \( G(7)=0 \)

Substitute \( x = 7 \) into \( G(x) \) and set it equal to 0:\[G(7) = \frac{1}{2} \cdot 7^2 - \frac{1}{2} \cdot a^2 = 0.\]This simplifies to:\[\frac{49}{2} = \frac{1}{2}a^2 \implies a^2 = 49 \implies a = 7.\]
05

Finalize the Function

Now we have \( a = 7 \), so the required function \( G(x) \) is:\[G(x) = \frac{1}{2}x^2 - \frac{49}{2}.\] This function is concave up, since the second derivative \( G''(x) = 1 > 0 \), and it satisfies \( G(7) = 0 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Concavity
Concavity in a function tells us about the shape of its graph.
When a function is concave up, it forms a U-like shape, meaning it curves upwards. This happens when the second derivative of the function is positive.
  • If a function's second derivative, denoted as \( G''(x) \), is greater than zero \( (G''(x) > 0) \), the graph is concave up.
  • If \( G''(x) < 0 \), then the graph is concave down, forming an upside-down U-like shape.
For a function \( G(x) = \int_a^x f(t) \, dt \) to be concave up, the function \( f(x) \) it derives from needs to have a positive derivative. In the original problem, the choice of \( f(x) = x \) ensures that \( G''(x) \) is always 1, thus making \( G(x) \) concave up.
Integration
Integration is like finding the area under a curve. When you integrate a function, you are essentially adding up small pieces of area to get a total area.
This process is fundamental in calculus for determining accumulated quantities.
  • The definite integral from \( a \) to \( x \) of \( f(t) \) is given as \( \int_a^x f(t) \, dt \).
  • The Second Fundamental Theorem of Calculus connects differentiation and integration by saying if you integrate a continuous function and then differentiate the result, you'll get the original function back, \( G'(x) = f(x) \).
When constructing \( G(x) \) in the exercise, integration was used to formulate \( G(x) = \frac{1}{2}x^2 - \frac{49}{2} \), showing the power of integration to define new functions.
Continuous Function
A continuous function has no gaps or breaks in its graph.
This means that small changes in \( x \) result in small changes in \( f(x) \).
  • For a function to be continuous on an interval \([a, b]\), it must be defined for every point in that interval, and show no jumps or interruptions.
  • Continuous functions are essential in calculus because they guarantee the applicability of important theorems, like the Second Fundamental Theorem of Calculus.
In the exercise solution, the function \( f(x) \) had to be continuous to apply the theorem correctly.
By ensuring \( f(x) = x \) is continuous, we could confidently use integration to construct the desired function \( G(x) \).
This is why verifying continuity is a crucial step in calculus exercises and mathematical analysis.

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Most popular questions from this chapter

Are the statements in Problems \(104-112\) true or false? Give an explanation for your answer. If \(F(x)\) is an antiderivative of \(f(x)\) and \(G(x)\) is an antiderivative of \(g(x),\) then \(F(x) \cdot G(x)\) is an antiderivative of \(f(x) \cdot g(x)\).

Find the exact positive value of \(c\) which makes the area under the graph of \(y=c\left(1-x^{2}\right)\) and above the \(x\) -axis equal to 1.

Are the statements in Problems \(104-112\) true or false? Give an explanation for your answer. If \(F(x)\) and \(G(x)\) are two antiderivatives of \(f(x)\) for \(-\infty < x < \infty\) and \(F(5) > G(5),\) then \(F(10) > G(10)\).

In Exercises \(6-21,\) find an antiderivative. $$g(t)=\sin t$$

In Exercises \(66-75,\) decide if the statement is True or False by differentiating the right-hand side. $$\int e^{2 x} d x=2 e^{2 x}+C$$
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