Problem 24 - Applied Calculus

Question

Use integration by parts to find each integral. \(\int \frac{x+1}{e^{3 x}} d x\)

Step-by-Step Solution

Answer

The integral evaluates to \(-\frac{1}{3}(x+1)e^{-3x} - \frac{1}{9}e^{-3x} + C\).

Step 1: Identify Parts

Integration by parts is based on the formula \( \int u \, dv = uv - \int v \, du \). We first identify the functions \( u \) and \( dv \) from our integral. Let's set \( u = x+1 \) and the remaining part as \( dv = \frac{1}{e^{3x}} \, dx = e^{-3x} \, dx \).

Step 2: Differentiate and Integrate

Now, differentiate \( u \) and find \( du \), and integrate \( dv \) to find \( v \).\[ du = d(x+1) = dx \]\[ v = \int e^{-3x} dx = -\frac{1}{3} e^{-3x} \]

Step 3: Substitute into Integration by Parts Formula

Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula.\[ \int (x+1) e^{-3x} \, dx = (x+1)(-\frac{1}{3} e^{-3x}) - \int (-\frac{1}{3} e^{-3x}) \, dx \]

Step 4: Simplify and Solve Remaining Integral

Simplify and solve the remaining integral.\[ -\frac{1}{3}(x+1)e^{-3x} + \frac{1}{3} \int e^{-3x} \, dx \]The remaining integral is:\[ \frac{1}{3} \int e^{-3x} \, dx = \frac{1}{3} \left(-\frac{1}{3} e^{-3x}\right) = -\frac{1}{9} e^{-3x} \]

Step 5: Combine Results

Combine the results from the integration by parts and the solved remaining integral. The integral is now given by:\[ -\frac{1}{3}(x+1)e^{-3x} - \frac{1}{9} e^{-3x} + C \]where \( C \) is the constant of integration.

Key Concepts

Definite Integrals

Definite integrals are a crucial concept in calculus that help us calculate the area under a curve over a specified range. Unlike indefinite integrals, which include a constant of integration, definite integrals have limits of integration and thus yield a specific numerical value. To compute a definite integral, we evaluate the indefinite integral at the upper and lower limits and subtract the results.

For example, in the function \( \int_a^b f(x) \, dx \), 'a' and 'b' are the limits. After finding the indefinite integral \( F(x) \), we calculate \( F(b) - F(a) \). This result provides the net area between the curve and the x-axis from \( x = a \) to \( x = b \).

This concept is widely used in different fields like physics and engineering to solve problems involving accumulation, like finding distances, areas, and even probabilities.

Integration Techniques

There are various integration techniques used to find integrals of different functions. One of the popular methods is "integration by parts", especially useful when dealing with the product of two functions. This technique is based on the integration by parts formula:

• \( \int u \, dv = uv - \int v \, du \)

To effectively use this method, it is important to choose the right functions for \( u \) and \( dv \). Usually, 'u' is chosen as the easily differentiable function, and \( dv \) becomes the remaining part of the integral that can be integrated.

Common strategies also involve choosing polynomial functions for 'u' and exponential or trigonometric functions for \( dv \). Once these parts are identified, you differentiate 'u' to find \( du \) and integrate \( dv \) to find 'v'. Substitute these into the formula and solve the remaining part, which is usually simpler than the original problem. With some practice, mastering these strategies can solve many complex calculus problems.

Calculus Problems

Solving calculus problems often requires understanding both differentiation and integration, the two main tools of calculus. Integration allows us to analyze rates of change in reverse, essentially "undoing" differentiation. This is invaluable across various applications, from evaluating areas under curves to solving real-world problems in physics and engineering.

Every calculus problem is unique, and one challenge is discerning which techniques and approaches will simplify the problem most efficiently. Problems can range from straightforward polynomial integrations to more complicated issues involving multiple methods or changes of variables.

By breaking a problem into smaller parts, perhaps simplifying an expression or using a combination of techniques like substitution and integration by parts, we can tackle the problem step by step. Problem-solving in calculus is as much about understanding the underlying concepts as it is about applying the correct formulas and methods. Developing a strong grasp of fundamental techniques and practicing regularly are key to success in calculus.