Problem 64 - Calculus for Biology and Medicine

Question

$$ \text { Find the value of } a \geq 0 \text { that maximizes } \int_{0}^{a}\left(4-x^{2}\right) d x \text { . } $$

Step-by-Step Solution

Answer

The value of \( a \) that maximizes the integral is 2.

Step 1: Understand the Problem

We are tasked to find the value of \( a \) that maximizes the definite integral \( \int_{0}^{a} (4-x^2) \, dx \). This means we need to find \( a \geq 0 \) for which the area under the curve \( 4-x^2 \) from \( x=0 \) to \( x=a \) is the largest.

Step 2: Set Up the Integral

The definite integral we need to evaluate is \( \int_{0}^{a} (4-x^2) \, dx \). This integral will give us the area under the curve from \( x=0 \) to \( x=a \).

Step 3: Integrate the Function

Compute the integral \( \int (4-x^2) \, dx \). Using basic integration rules:\[\int (4-x^2) \, dx = 4x - \frac{x^3}{3} + C\]Evaluate the definite integral by applying the limits of integration from 0 to \( a \).

Step 4: Evaluate the Definite Integral

Substitute the limits into the integrated function:\[\left[4x - \frac{x^3}{3}\right]_0^a = \left(4a - \frac{a^3}{3}\right) - \left(4 \times 0 - \frac{0^3}{3}\right)\]Simplifying gives:\[4a - \frac{a^3}{3}\]

Step 5: Differentiate to Find Critical Points

To find the value that maximizes the integral, take the derivative with respect to \( a \):\[d\left(4a - \frac{a^3}{3}\right) / da = 4 - a^2\]Set this derivative equal to zero to find the critical points:\[4 - a^2 = 0\]

Step 6: Solve for a

Solve \( 4 - a^2 = 0 \):\[a^2 = 4\]\[a = 2 \quad \text{(since } a \geq 0\text{)}\]

Step 7: Verify Maximality

To ensure this value of \( a \) gives a maximum, check the second derivative:\[d^2\left(4a - \frac{a^3}{3}\right) / da^2 = -2a\]Evaluate at \( a = 2 \):\(-2 \times 2 = -4 < 0\).Since the second derivative is negative, \( a = 2 \) is indeed a maximum.

Key Concepts

Definite Integral

The concept of a definite integral is foundational in calculus. It represents the accumulation of quantities, such as areas, over an interval. Simply put, a definite integral calculates the total area under a curve over a specified range. In our problem, we have the function

• \( f(x) = 4 - x^2 \)

We want to find the definite integral from \( x = 0 \) to \( x = a \). This is written as

• \( \int_{0}^{a} (4-x^2) \, dx \).

The result from the integration process gives the expression of the area under the curve from \( x = 0 \) to \( x = a \). In practical terms, evaluating this integral helps in determining how the area changes as \( a \) varies. Once integrated, the expression becomes

• \( 4a - \frac{a^3}{3} \).

This expression represents the total area under the curve from the start point \( x = 0 \) to the end point \( x = a \). Understanding definite integrals is crucial since they allow us to analyze and optimize such areas.

Critical Points

Identifying critical points is an essential step in optimization problems. Critically, these points are where the first derivative of a function equals zero, indicating potential maxima or minima. After obtaining the expression of the area under the curve

• \( 4a - \frac{a^3}{3} \),

we need to find the critical points by differentiating it with respect to \( a \).

• The first derivative is \( \frac{d}{da} \left( 4a - \frac{a^3}{3} \right) = 4 - a^2 \).

Setting this derivative to zero helps locate points of potential interest:

• \( 4 - a^2 = 0 \)

Solving for \( a \) gives

• \( a = 2 \).

This is the critical point because at this value of \( a \), the rate of change of the area becomes zero. Examining critical points gives insight into the extremal values of the function, essential for finding a local maximum or minimum in the context of optimization.

Second Derivative Test

The second derivative test is used to confirm the nature of critical points, whether they are maxima, minima, or points of inflection. After finding a critical point by setting the first derivative to zero, evaluating the second derivative at that point helps determine if it's a maximum or minimum.

• The second derivative of the function \( 4a - \frac{a^3}{3} \) with respect to \( a \) is \( -2a \).

Substituting the critical point \( a = 2 \) into the second derivative yields

• \( -2 \times 2 = -4 \).

A negative result indicates a local maximum. Thus, the second derivative test confirms that at \( a = 2 \), the original function \( 4a - \frac{a^3}{3} \) achieves a maximum value. In essence, the second derivative test provides a reliable way to validate whether a critical point represents a peak or a trough in the context of a graph, crucial for optimization problems where maximum or minimum values are sought.