Problem 64 - Applied Calculus

Question

Find \(\int(x+1) d x\) : a. By using the formula for \(\int u^{n} d u\) with \(n=1\). b. By dropping the parentheses and integrating directly. c. Can you reconcile the two seemingly different answers? [Hint: Think of the arbitrary constant.]

Step-by-Step Solution

Answer

Both methods yield \( \frac{x^2}{2} + x + C \), consistent with each other when considering the arbitrary constant \( C \).

Step 1: Applying the Power Rule for Integration

To find \( \int(x+1) dx \) using the formula \( \int u^n du \), we recognize \( n=1 \). Therefore, the integral of \( x^1 \) is \( \frac{x^{1+1}}{1+1} + C \), where \( C \) is the integration constant. Calculating this gives us \( \frac{x^2}{2} + C \). Similarly, integrating \( 1 \) with respect to \( x \) gives \( x + C \). The combined result is \[ \int(x+1) dx = \frac{x^2}{2} + x + C. \]

Step 2: Integrating by Dropping the Parentheses

When integrating \( \int(x+1) dx \), we can think of it as two separate integrals: \( \int x \, dx \) and \( \int 1 \, dx \). Using the power rule, the integral of \( x \) is \( \frac{x^2}{2} \) and the integral of the constant \( 1 \) is \( x \). Therefore, the result is \[ \int(x+1) dx = \frac{x^2}{2} + x + C. \]

Step 3: Reconciling the Two Approaches

Both methods, the formula for \( \int u^n du \) and direct integration by dropping parentheses, yield the same result: \( \frac{x^2}{2} + x + C \). The apparent difference lies in the constant \( C \), which is arbitrary and represents any constant number. Therefore, any difference can be accounted for by adjusting \( C \).

Key Concepts

Power Rule

The Power Rule is a fundamental concept in calculus that simplifies the process of finding the integral, especially when dealing with polynomial expressions. It states that the integral of a power of a variable is given by the formula:

• \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \) , where \( n eq -1 \).

In the exercise, we first recognize that we have a polynomial term, \( x^1 \), within \( \int(x+1) \, dx \). Using the power rule here, you add one to the exponent, then divide by the new exponent:

• \( \int x^1 \, dx = \frac{x^{1+1}}{1+1} = \frac{x^2}{2} \).

The power rule makes evaluating these terms straightforward and ensures consistent results during integration.
Understanding this rule is crucial as it forms a backbone for handling more complex integration problems.

Integration Constant

When integrating functions, especially indefinite integrals, you may notice an unknown constant, represented by \( C \). This constant is crucial because integration represents the process of finding the antiderivative. The antiderivative, unlike derivatives, isn't unique. There can be infinitely many functions differing by a constant, all of which have the same derivative.

• For example, \( rac{x^2}{2} + x + C \) can be such an antiderivative for \( x+1 \).

Consider that derivatives remove constants, but integration reintroduces them. Hence, every result of indefinite integration includes \( C \), ensuring no lost information.
In problems, different approaches may lead to different forms of \( C \), but they all reflect the same family of functions.

Definite Integrals

While the discussed exercise focuses on indefinite integrals, it's good to understand definite integrals as they contrast interestingly. Definite integrals calculate the net area between a function and the x-axis over a specific interval.

• Represented by: \( \int_a^b f(x) \, dx \).
• It yields a numerical result, unlike the family of functions from indefinite integrals.

Since definite integrals involve limits \( a \) and \( b \), the integration constant \( C \) cancels itself out. Calculating a definite integral provides exact quantities, often used in real-world physical applications such as finding distances or volumes. Understanding both types enriches your grasp on how calculus applies differently in various contexts.

Indefinite Integrals

Indefinite integrals refer to finding the antiderivative of a function without specific limits, like in the given exercise. They express functions generally and include the integration constant \( C \):

• The format is: \( \int f(x) \, dx = F(x) + C \).
• This family of functions encompasses all potential solutions differing by constants.

Using indefinite integrals, we can reconstruct functions and solve differential equations without boundary specifics. They often serve as intermediate steps in solving definite integrals, where the specific range is then applied.
If you're looking to solve integrals broadly, investigating indefinite integrals is your route to a versatile toolbox for further mathematical exploration.