Problem 43 - Calculus for Biology and Medicine

Question

As definite integrals. Note that (1) \(P=\left[x_{0}, x_{1}, \ldots, x_{n}\right]\) is a partition of the indicated interval, (2) \(c_{k} \in\left[x_{k-1}, x_{k}\right]\), and (3) \(\Delta x_{k}=x_{k}-x_{k-1}\). \(\lim _{\mid P \| \rightarrow 0} \sum_{k=1}^{n}\left(2 c_{k}-1\right) \Delta x_{k}\), where \(P\) is a partition of \([-3,2]\)

Step-by-Step Solution

Answer

The value of the integral is \(-10\).

Step 1: Identify the Function

The limit of the Riemann sum is given as \( \lim _{\mid P \| \rightarrow 0} \sum_{k=1}^{n}(2 c_{k}-1) \Delta x_{k} \). Here, the function being integrated is represented by \( f(x) = 2x - 1 \).

Step 2: Determine the Interval

The integration is performed over the interval provided in the partition \([-3, 2]\). As \(x_0 = -3\) and \(x_n = 2\), this will be the interval for the definite integral of \(f(x)\).

Step 3: Define the Definite Integral

Convert the Riemann sum into a definite integral over the interval \([-3, 2]\). The integral is \( \int_{-3}^{2} (2x - 1) \, dx \).

Step 4: Compute the Definite Integral

Calculate the integral \( \int (2x - 1) \, dx \). The antiderivative of \(2x\) is \(x^2\), and the antiderivative of \(-1\) is \(-x\). Thus, the antiderivative of the entire function is \(x^2 - x\).

Step 5: Evaluate the Integral

Use the Fundamental Theorem of Calculus to evaluate the definite integral: \[ \int_{-3}^{2} (2x - 1) \, dx = \left[ x^2 - x \right]_{-3}^{2} \]. Compute \((2)^2 - 2 - ((-3)^2 + 3)\), which results in \[ (4 - 2) - (9 + 3) = 2 - 12 = -10 \].

Key Concepts

Riemann sum

The Riemann sum is an essential concept in calculus because it helps us understand integrals as limits of sums. A Riemann sum approximates the area under a curve by summing up areas of simple shapes like rectangles. Consider a function, say, \( f(x) = 2x - 1 \), which you'd like to integrate over a certain interval, for instance, \([-3, 2]\). To find the total area under the curve from \( x = -3 \) to \( x = 2 \), you break the interval into smaller partitions, called \( P = [x_0, x_1, \ldots, x_n] \).
These partitions have sub-interval lengths of \( \Delta x_k = x_k - x_{k-1} \). At each sub-interval, you evaluate the function \( f(x) \) at a chosen point, \( c_k \), which lies within each small interval \([x_{k-1}, x_k]\).
Thus, a Riemann sum can be represented as \( \sum_{k=1}^{n} (f(c_k)) \Delta x_k \). The more sub-intervals you use, the more accurate your estimate. Eventually, if the width of the sub-intervals goes to zero \( (||P|| \rightarrow 0) \), this sum becomes the integral of the function over the interval.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation with integration and provides a powerful tool for evaluating definite integrals. This theorem has two parts.

• The first part states that, if a function is integrable on an interval, an antiderivative of that function can be used to find the definite integral over that interval.
• The second part tells us that if you know the antiderivative of a function, you can evaluate the integral over an interval using the difference of values of the antiderivative at the endpoints of the interval.

In the context of our initial problem, the Fundamental Theorem of Calculus helped us evaluate the definite integral \( \int_{-3}^{2} (2x - 1) \, dx \).
First, we found the antiderivative, \( F(x) = x^2 - x \), and then applied the theorem by computing \( F(2) - F(-3) \), leading to the result of \(-10\).
This makes evaluating integrals much more straightforward than summing infinite partitions by Riemann sums.

antiderivative

Finding an antiderivative is a crucial step in solving integrals. An antiderivative of a function \( f(x) \) is another function \( F(x) \) whose derivative is \( f(x) \). In simpler terms, if you differentiate \( F(x) \), you should get back \( f(x) \).
For example, consider the function \( f(x) = 2x - 1 \). To find its antiderivative, you'll need to reverse the derivative operations:

• The antiderivative of \( 2x \) is \( x^2 \). To reach this, note that the derivative of \( x^2 \) is \( 2x \).
• For constant terms, such as \(-1\), the antiderivative is \(-x\), since the derivative of \(-x\) is \(-1\).

When we combine these, the antiderivative for the entire function becomes \( x^2 - x \).
Remember, the constant of integration is often omitted in definite integrals because it cancels out when evaluating the limit of integration. Understanding antiderivatives makes solving calculus problems less intimidating, especially when combined with the Fundamental Theorem of Calculus.