Problem 38 - Biocalculus Calculus for the Life Sciences

Question

Find the general indefinite integral. Illustrate by graphing several members of the family on the same screen. \(\int\left(e^{x}-2 x^{2}\right) d x\)

Step-by-Step Solution

Answer

The general integral is \( e^x - \frac{2x^3}{3} + C \).

Step 1: Identify the Integral

The integral we need to solve is \( \int (e^x - 2x^2) \, dx \). This means we have to find the antiderivative of the function \( e^x - 2x^2 \).

Step 2: Apply Antiderivative Rules

To integrate \( e^x - 2x^2 \), we apply the rule \( \int e^x \, dx = e^x + C \) for the exponential function, and \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) for polynomials. Thus, \( \int -2x^2 \, dx = -\frac{2x^3}{3} + C \). Combine these to form the antiderivative.

Step 3: Calculate the Antiderivative

Apply the rules: \( \int e^x \, dx = e^x + C_1 \) and \( \int -2x^2 \, dx = -\frac{2x^3}{3} + C_2 \). Combining these, the general antiderivative is \( e^x - \frac{2x^3}{3} + C \) where \( C = C_1 + C_2 \) is the constant of integration.

Step 4: Graph Multiple Functions

To illustrate, we graph several members of the family of functions by using different values for the constant \( C \). For example, use \( C = 0, 1, -1 \) to plot \( e^x - \frac{2x^3}{3} + 0 \), \( e^x - \frac{2x^3}{3} + 1 \), and \( e^x - \frac{2x^3}{3} - 1 \). Each graph should look similar but be shifted vertically by the value of \( C \).

Key Concepts

Antiderivative Rules

When tackling integrals, which is essentially the reverse process of differentiation, one must be familiar with antiderivative rules. These are critical tools used in calculus to determine the original function from its derivative.
For the function given, the integral \( \int(e^x - 2x^2) \ dx \), we look at the antiderivative rules:

• The rule for exponential functions: \( \int e^x \, dx = e^x + C \), where \( C \) is the constant of integration.
• For polynomials, the rule is \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), a formula crucial while analyzing terms like \( -2x^2 \). Hence, \( \int -2x^2 \, dx = -\frac{2x^3}{3} + C \).

By applying these rules, we can successfully find the general antiderivative of the combined functions, recognizing that each solution will include a constant \( C \).
Understanding these rules allows us to tackle a wide range of problems involving indefinite integrals of various forms of functions.

Exponential Functions

Exponential functions are a staple in both mathematics and real-world applications. They follow the form \( f(x) = e^x \), where \( e \) is Euler's number, approximately 2.718. The key feature of exponential functions is their growth behavior, which increases rapidly.
The integral of an exponential function, such as from our example, \( \int e^x \, dx \,\), is \( e^x + C \). This is due to the unique property of exponential functions, where they are their own derivatives.

• Exponentials naturally accommodate growth and decay problems, making them valuable in science and finance.
• For integration, simplicity is offered by the fact that the integral keeps the same form as the original function, only adding the constant \( C \) to account for any shifts.

When plotting graphically, different constants will provide vertically shifted versions of the function \( e^x - \frac{2x^3}{3} + C \), maintaining the exponential growth beyond the polynomial term's influence.

Polynomial Integration

In polynomial integration, we handle terms of the form \( x^n \) to find the antiderivative, a core skill in calculus. For instance, in the function \( -2x^2 \,\) part of \( \int e^x - 2x^2 \, dx \,\), we use the rule \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) to achieve \( \int -2x^2 \, dx = -\frac{2x^3}{3} + C \).

Here's how polynomial integration functions:

• A constant, like \( -2 \), scales the polynomial's antiderivative.
• Adding \( 1 \) to the original exponent \( n \,\) and then dividing by the new exponent \( n+1 \,\) balances the term.

This approach then blends with exponential integrations to solve complex equations.
The polynomials' outcome, as visible when graphing, affects the curve differently than exponential terms, providing smooth, elegant curves as each integral shifts corresponds to different constants \( C \.\)