Problem 81 - Biocalculus Calculus for the Life Sciences

Question

Suppose \(h\) is a function such that \(h(1)=-2, h^{\prime}(1)=2\) \(h^{\prime \prime}(1)=3, h(2)=6, h^{\prime}(2)=5, h^{\prime \prime}(2)=13,\) and \(h^{\prime \prime}\) is continuous everywhere. Evaluate \(\int_{1}^{2} h^{\prime \prime}(u) d u\)

Step-by-Step Solution

Answer

The integral \( \int_{1}^{2} h^{\prime \prime}(u) \, du \) is 3.

Step 1: Understand the Problem

We are given values of a function, its first derivative, and its second derivative at two points. We are asked to evaluate the integral of the second derivative from 1 to 2, i.e., \( \int_{1}^{2} h^{\prime \prime}(u) \, du \).

Step 2: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus tells us that the integral of the derivative of a function over an interval gives us the change in the function values over that interval. In this case \( \int_{1}^{2} h^{\prime \prime}(u) \, du = h^{\prime}(2) - h^{\prime}(1) \).

Step 3: Substitute Given Values

From the problem, we know that \( h^{\prime}(2) = 5 \) and \( h^{\prime}(1) = 2 \). Substituting these values into our equation gives us \( \int_{1}^{2} h^{\prime \prime}(u) \, du = 5 - 2 \).

Step 4: Calculate the Result

Now perform the subtraction: \( 5 - 2 = 3 \). This means that the value of \( \int_{1}^{2} h^{\prime \prime}(u) \, du \) is 3.

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration, two main operations in calculus. It comprises two main parts, with the first part connecting a function with its antiderivative and the second, often more used, linking the definite integral of a function to its antiderivatives.

Here is the crucial takeaway: If you have a function that's integrable on a certain interval, say \[a, b\], the Fundamental Theorem of Calculus tells us the definite integral of its derivative over \[a, b\] can be calculated by evaluating the change in the original function across those endpoints. Mathematically, if \( F \) is the antiderivative of \( f \), it says:

• \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \)

This theorem simplifies the evaluation of definite integrals quite significantly. In the exercise, we used this theorem to compute \( \int_{1}^{2} h^{\prime \prime}(u) \, du \) by directly subtracting the values of the first derivative of \( h \) at the points 2 and 1.

Derivatives

Derivatives measure how a function changes as its input changes. They tell us the slope of a function at any given point and are crucial in understanding the rate at which quantities change. In formal terms, the derivative of a function \( f \) at a point \( x \) is given by:

• \( f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)

We primarily encounter first derivatives, which reflect the rate of change of the function, but second derivatives are also critical. They show how the rate of change of the function is itself changing. For example, in physics, the second derivative of a position function gives acceleration, which is the rate of change of velocity (the first derivative).

In our exercise, \( h^{\prime \prime}(x) \) represents the second derivative of \( h \), and it provides valuable information about the curvature of the function \( h \). The integral of this second derivative over a range helps in understanding the overall change in the first derivative, which is essential in numerous applications, from physics to economics.

Continuity

A function is said to be continuous when there are no breaks, jumps, or holes in its graph. More formally, a function \( f \) is continuous at a point \( c \) if:

• \( \lim_{x \to c} f(x) = f(c) \)

Continuity ensures that the behavior of the function is predictable and smooth from one value to the next. When it comes to derivatives, if a function's derivative is continuous, it means that the function changes at a consistent rate.

In the exercise, it's given that \( h^{\prime \prime} \) is continuous everywhere. This is important because it assures us that when we integrate \( h^{\prime \prime} \), the result will be a well-behaved, smooth function. Continuity, in this context, simplifies working with integrals and ensures there are no complications like undefined behavior.

• Ensures smooth transition between values
• Simplifies integration and differentiation processes
• Leads to predictable and reliable mathematical behavior