Problem 6 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Approximate A. \(\int_{0}^{1} e^{t} d t\) B. \(\int_{0}^{\pi} \sin (t) d t\)

Step-by-Step Solution

Answer

A: Approximately 1.71828 B: Exactly 2

Step 1: Understanding the Problem

We are tasked with approximating two definite integrals: 1. \( \int_{0}^{1} e^{t} dt \)2. \( \int_{0}^{\pi} \sin(t) dt \)We will use the properties of definite integrals and the Fundamental Theorem of Calculus to find these integrals.

Step 2: Recall the Antiderivative

To solve these integrals, we first need to know the antiderivatives (indefinite integrals) of the functions involved.- For \( e^t \), the antiderivative is still \( e^t \).- For \( \sin(t) \), the antiderivative is \( -\cos(t) \).

Step 3: Apply the Fundamental Theorem of Calculus to Part A

The Fundamental Theorem of Calculus states:\[ \int_{a}^{b} f(t) \, dt = F(b) - F(a) \]where \( F \) is an antiderivative of \( f \).For \( \int_{0}^{1} e^t dt \): - Antiderivative is \( e^t \).- Evaluate as: \( e^1 - e^0 = e - 1 \).

Step 4: Evaluate Numerical Result for Part A

Using the approximate value of \( e \approx 2.71828 \), \( e - 1 \approx 2.71828 - 1 = 1.71828 \).

Step 5: Apply the Fundamental Theorem of Calculus to Part B

For \( \int_{0}^{\pi} \sin(t) dt \): - Antiderivative is \( -\cos(t) \).- Evaluate as: \( (-\cos(\pi)) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2 \).

Step 6: Summary of Results

After evaluating both integrals, we have:1. \( \int_{0}^{1} e^t dt \approx 1.71828 \)2. \( \int_{0}^{\pi} \sin(t) dt = 2 \)

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is an essential concept in finding definite integrals. It connects differentiation and integration, two central operations in calculus. The theorem consists of two parts. The first part provides an antiderivative of a function if the function is integrable on a closed interval. The second part allows one to evaluate the definite integral of a function if an antiderivative is known. In other words, if you have a function \( f(x) \) that is continuous over an interval \([a, b]\), and \( F(x) \) is an antiderivative of \( f(x) \), then the definite integral from \( a \) to \( b \) of \( f(x) \) is given by:

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

By using this formula, we can bypass the detailed calculation of areas under curves. Instead, we focus on finding antiderivatives – a much simpler task in many cases. This foundational theorem streamlines problem-solving, making it a powerful tool in calculus.

Antiderivative

An antiderivative, also known as an indefinite integral, is a function that reverses the process of differentiation. If a function \( F(x) \) is an antiderivative of \( f(x) \), then \( F'(x) = f(x) \). Essentially, finding an antiderivative is achieving the inverse of taking a derivative. Knowing how to find antiderivatives is crucial for solving definite integrals with the help of the Fundamental Theorem of Calculus.
For example, in the case of the functions in the exercise:

• The antiderivative of \( e^t \) is \( e^t \).

• The antiderivative of \( \sin(t) \) is \( -\cos(t) \).

With these antiderivatives, we can easily apply them to evaluate the definite integrals by plugging the boundary values into the antiderivative function. Remember, the calculations become straightforward once we identify the correct antiderivative.

Numerical Approximation

Numerical approximation is the process of finding a non-exact, yet close solution for integrals and other mathematical calculations. Sometimes, finding the exact value of an integral can be challenging or impossible analytically, especially for complex functions. This is where numerical methods come in. Approximations like the ones used in the exercise help make calculations manageable and are often sufficient for practical purposes.
In our case, we approximated \( e - 1 \) by using the common approximation \( e \approx 2.71828 \), resulting in \( 1.71828 \) for

• \( \int_{0}^{1} e^t dt \).

Some common numerical approximation techniques include:

• Trapezoidal Rule
• Simpson's Rule
• Riemann Sums

These methods divide the interval into small sections and compute approximate values to achieve results that approach the true value of the integral. Grasping numerical approximations makes it easier to tackle real-world applications where precision is essential yet perfect accuracy may not be feasible.