Problem 10 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

$$ \int_{0}^{1} \frac{x+2}{e^{3 x}} d x $$

Step-by-Step Solution

Answer

\( \int_{0}^{1} \frac{x+2}{e^{3 x}} d x = \frac{7}{9} - \frac{1}{e^{3}} \)

Step 1: Rewrite the Integral

Separate the integral into two parts to make it easier to solve: \[ \frac{x + 2}{e^{3x}} = \frac{x}{e^{3x}} + \frac{2}{e^{3x}} \] Now, we can rewrite the integral as: \[ \begin{aligned} \ \ \ \ \ \ \ \ \ \ \int_{0}^{1} \frac{x+2}{e^{3 x}} d x = \int_{0}^{1} \frac{x}{e^{3 x}} d x + \int_{0}^{1} \frac{2}{e^{3 x}} d x \end{aligned} \]

Step 2: Solve the First Integral

Use integration by parts for the first integral, where \( u = x \) and \( dv = e^{-3x} \, dx \). Then calculate \( du = dx \) and \( v = -\frac{1}{3}e^{-3x} \): \[ \int_{0}^{1} \frac{x}{e^{3 x}} d x = \int_{0}^{1} x e^{-3x} d x \] Using integration by parts formula \( \int u \, dv = uv - \int v \, du \): \[ \begin{aligned} \ \ \ &= \left. -\frac{x}{3} e^{-3 x} \right]_{0}^{1} + \int_{0}^{1} \frac{1}{3} e^{-3 x} d x \ &= \left. -\frac{x}{3} e^{-3 x} \right]_{0}^{1} + \left. -\frac{1}{9} e^{-3 x} \right]_{0}^{1} \ &= \left( -\frac{1}{3} e^{-3} + 0 \right) + \left( -\frac{1}{9} e^{-3} + \frac{1}{9} \right) \ &= -\frac{1}{3e^{3}} - \frac{1}{9e^{3}} + \frac{1}{9} \end{aligned} \]

Step 3: Solve the Second Integral

For the second integral, use a basic integration rule. We have:\[ \int_{0}^{1} \frac{2}{e^{3 x}} d x = 2 \int_{0}^{1} e^{-3 x} d x \] The integral of \( e^{-3x} \) is \( -\frac{1}{3}e^{-3x} \), so we get:\[ \left.2 \cdot -\frac{1}{3} e^{-3 x} \right]_{0}^{1} = 2 \cdot -\frac{1}{3} [e^{-3} - 1] = -\frac{2}{3} e^{-3} + \frac{2}{3} \]

Step 4: Combine the Results

Add the results from the two integrals:\[ \begin{aligned} \ \ \int_{0}^{1} \frac{x+2}{e^{3x}} d x &= \left( -\frac{1}{3e^{3}} - \frac{1}{9e^{3}} + \frac{1}{9} \right) + \left( -\frac{2}{3e^{3}} + \frac{2}{3} \right) \ &= -\frac{1}{3e^{3}} - \frac{1}{9e^{3}} - \frac{2}{3e^{3}} + \left( \frac{1}{9} + \frac{2}{3} \right) \ &= -\left( \frac{3}{9e^{3}} + \frac{1}{9e^{3}} + \frac{6}{9e^{3}} \right) + \left( \frac{1}{9} + \frac{6}{9} \right) \ &= -\frac{1}{e^{3}} + \frac{7}{9} \ &= \frac{7}{9} - \frac{1}{e^{3}} \end{aligned} \]

Key Concepts

Integration by Parts

Understanding integration by parts is essential for tackling integrals involving the product of functions.
It's based on the product rule for differentiation. Here’s a rundown of how it works:
Given two functions, say, \(u(x)\) and \(v(x)\), the integration by parts formula is:
\[ \int u \, dv = uv - \int v \, du \]
Suppose we solve an integral like \[ \int x e^{-3x} \, dx \]. We select:

• \(u = x\) (which simplifies when differentiated).
• \(dv = e^{-3x} \, dx\) (easy to integrate).

Then, differentiate \(u\) to get \(du = dx\), and integrate \(dv\) to get \(v = -\frac{1}{3}e^{-3x}\).
Substituting into the formula:
\[ uv - \int v \, du \]
We get:
\[ -\frac{x}{3}e^{-3x} \bigg|_{0}^{1} + \int_{0}^{1} \frac{1}{3}e^{-3x} \, dx \]
This method simplifies complex integrals into manageable terms.

Exponential Functions

Exponential functions are a fundamental concept in calculus.
They take the form \(f(x) = a^x\), where \(a\) is a positive constant. A common type is the natural exponential function, given by \(e^x\).
This function has key properties:

• The derivative of \(e^x\) is \(e^x\), making it unique since the function and its derivative are the same.
• The integral of \(e^{kx}\), where \(k\) is a constant, is \(\frac{1}{k}e^{kx}\).

For example, integrating \(e^{-3x}\) yields \(\frac{1}{-3}e^{-3x}\).
These functions are integral to many calculus problems due to their distinctive rate of growth and decay characteristics.

Basic Integration Rules

Basic integration rules form the cornerstone of solving integral problems.
Some of the most frequently used rules include:

• Power Rule: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \, \text{where } n eq -1 \)
• Constant Multiple Rule: \(\int k f(x) \, dx = k \int f(x) \, dx \)
• Sum Rule: \( \int (f(x) + g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx \)

When solving integrals such as \( \int_{0}^{1} \frac{2}{e^{3x}} \, dx \), we use the Constant Multiple Rule:
Since \(\frac{2}{e^{3x}} = 2 e^{-3x}\), it simplifies our work:
\[ \int_{0}^{1} \frac{2}{e^{3x}} \, dx = 2 \int_{0}^{1} e^{-3x} \, dx \]
Recognizing and applying these basic rules makes complex integration problems much simpler to solve.