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Magazine Chapter 6: Problem 80
Find the value of \(a \geq 0\) that maximizes \(\int_{0}^{a}\left(4-x^{2}\right) d x\).
Short Answer
Expert verified
The value of \(a\) that maximizes the integral is \(a = 2\).
Step by step solution
01
Understand the Problem
We need to find the value of \(a\) that maximizes the definite integral \(\int_{0}^{a}\left(4-x^{2}\right)dx\). The expression inside the integral, \(4-x^2\), defines a curve from which we need to find the area between \(x = 0\) and \(x = a\).
02
Find the Antiderivative
The first step in solving the integral \(\int_{0}^{a}\left(4-x^{2}\right)dx\) is to determine an antiderivative of the integrand. The antiderivative of \(4-x^2\) is \(4x - \frac{x^3}{3}\).
03
Evaluate the Definite Integral
Now, evaluate the integral from 0 to \(a\) using the antiderivative found in Step 2. Apply the Fundamental Theorem of Calculus:\[\int_{0}^{a}\left(4-x^{2}\right)dx = \left[4x - \frac{x^3}{3}\right]_0^a = \left(4a - \frac{a^3}{3}\right) - (4 \cdot 0 - \frac{0^3}{3}) = 4a - \frac{a^3}{3}.\]
04
Determine the Critical Points
To find the value of \(a\) that maximizes the integral, differentiate \(4a - \frac{a^3}{3}\) with respect to \(a\):\[\frac{d}{da}\left(4a - \frac{a^3}{3}\right) = 4 - a^2.\]Set the derivative equal to zero and solve for \(a\):\[4 - a^2 = 0\] \[a^2 = 4\] \[a = 2\] (since \(a \geq 0\)).
05
Verify Maximum Value
Verify that \(a = 2\) is a maximum by using the second derivative test. Calculate the second derivative:\[\frac{d^2}{da^2}\left(4a - \frac{a^3}{3}\right) = -2a.\]Evaluate the second derivative at \(a = 2\):\[-2(2) = -4\]Since the second derivative is negative, \(a = 2\) is a point of local maximum.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
A definite integral is an important concept in calculus that represents the total accumulation of a quantity. Specifically, it calculates the net area under the curve of a function from one point to another on the x-axis. In the given exercise, the definite integral \( \int_{0}^{a}\left(4-x^{2}\right)dx \) helps find the total area from \( x = 0 \) to \( x = a \). This area is bounded by the curve \( 4 - x^2 \), the x-axis, and the vertical lines at \( x = 0 \) and \( x = a \).
- The curve \( 4 - x^2 \) is a downward-facing parabola, touching the y-axis at 4.
- Finding this area involves using the results from the antiderivative.
Antiderivative
The antiderivative is the reverse process of differentiation. It is a function whose derivative produces the original function. For the problem at hand, the antiderivative of \( 4 - x^2 \) is \( 4x - \frac{x^3}{3} \).
- From here, we use the antiderivative to evaluate the definite integral.
- Basically, this step replaces the process of summing up all small pieces with a straightforward calculation using the fundamental theorem of calculus.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a pivotal theorem that links differentiation to integration. It tells us two main things:
The theorem makes it easier to handle definite integrals by shifting the focus from summation to finding function values, a significant simplification in calculus.
- Part 1 states that if a function is continuous over an interval, then integrating it will produce an antiderivative.
- Part 2, which applies here, tells us how to evaluate a definite integral using an antiderivative.
The theorem makes it easier to handle definite integrals by shifting the focus from summation to finding function values, a significant simplification in calculus.
Second Derivative Test
The second derivative test is a technique used to determine whether a critical point found in a function is a maximum, minimum, or neither. After differentiating \( 4a - \frac{a^3}{3} \) to find the critical points, we get \( a = 2 \). The next step is to use the second derivative test to confirm if it is a point of maximum or minimum value.
Therefore, the value of \( a \) that maximizes the integral is accurately found using this test, which is an essential process in optimization problems within calculus.
- The second derivative of \( 4a - \frac{a^3}{3} \) is \( -2a \).
- Substituting \( a = 2 \) gives \( -4 \), which is negative.
Therefore, the value of \( a \) that maximizes the integral is accurately found using this test, which is an essential process in optimization problems within calculus.
