Problem 263 - Calculus Volume 1

Question

In the following exercises, find the antiderivative using the indicated substitution. $$ \int(2 x-3)^{-7} d x ; u=2 x-3 $$

Step-by-Step Solution

Answer

The antiderivative is \(-\frac{1}{12(2x-3)^6} + C\).

Step 1: Identify substitution

We are given the substitution \( u = 2x - 3 \). This substitution will simplify the integral by replacing \( 2x - 3 \) with \( u \).

Step 2: Differentiate substitution

Differentiate the substitution equation \( u = 2x - 3 \) with respect to \( x \) to find \( \frac{du}{dx} \). \[ \frac{du}{dx} = 2 \] Thus, we have \( du = 2 dx \), or \( dx = \frac{1}{2} du \).

Step 3: Substitute variables in the integral

Replace \( 2x - 3 \) with \( u \) and \( dx \) with \( \frac{1}{2} du \) in the integral: \[ \int (2x-3)^{-7} \, dx = \int u^{-7} \cdot \frac{1}{2} \, du \] which simplifies to \[ \frac{1}{2} \int u^{-7} \, du \].

Step 4: Find the antiderivative

To find the antiderivative of \( u^{-7} \), we use the power rule for integration: \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \) where \( n eq -1 \). Here \( n = -7 \), so \( n+1 = -6 \). The antiderivative is: \[ \frac{1}{2} \cdot \frac{u^{-6}}{-6} + C \].

Step 5: Simplify the expression

Simplify the expression obtained in the previous step: \[ -\frac{1}{12} u^{-6} + C \] which is equivalent to \[ -\frac{1}{12} \cdot \frac{1}{u^6} + C \].

Step 6: Substitute back in terms of x

Replace \( u \) with our original substitution, \( 2x - 3 \), to express the antiderivative in terms of \( x \): \[ -\frac{1}{12} \cdot \frac{1}{(2x-3)^6} + C \].

Key Concepts

Integration by Substitution

Integration by substitution is a fundamental technique in calculus that simplifies finding the antiderivative of complex functions. It is analogous to the chain rule in differentiation, allowing us to reverse the process. In our exercise, the substitution given is \( u = 2x - 3 \).

Here's a step-by-step guide to better understand this:

• **Identify the substitution**: By setting \( u = 2x - 3 \), we simplify the function inside the integral.
• **Differentiate the substitution**: Find \( \frac{du}{dx} \) to understand how \( u \) relates to \( x \). This helps convert \( dx \) to a function of \( du \).
• **Replace in the integral**: Substitute \( (2x-3) \) with \( u \) and \( dx \) with \( \frac{1}{2}du \) in your integral to make it easier to integrate.

By following these steps, integration becomes more straightforward, leading us to a simpler problem that's easier to solve.

Power Rule for Integration

The power rule for integration is another essential tool in calculus. It helps us find antiderivatives of functions in the form \( u^n \). The general formula is:
\[\int u^n \, du = \frac{u^{n+1}}{n+1} + C\]where \( n eq -1 \).

In our exercise, we encounter the integral \( \int u^{-7} \, du \). Applying the power rule:

• Identify \( n = -7 \).
• Compute \( n + 1 = -6 \).
• Substitute into the formula to find: \( \frac{u^{-6}}{-6} \).

Finally, remember to simplify your expression and don't forget the constant of integration \( C \), which represents the family of antiderivative functions.

Calculus Exercises

Working through calculus exercises can bolster your understanding of integration techniques like substitution and the power rule. Exercises such as the one provided encourage application of theoretical concepts to practical problems.

Here are some tips to tackle these types of exercises:

• Understand the problem by carefully reading the given information and directions.
• Break down the problem into smaller parts, such as identifying substitutions and differentiating them.
• Simplify as much as possible to make integration easier.
• Always substitute back in terms of the original variable at the end.

Incorporating these strategies ensures clearer understanding and greater success in solving calculus problems.