Problem 57 - Thomas Calculus

Question

Verify the formulas in Exercises 57-62 by differentiation. $$\int(7 x-2)^{3} d x=\frac{(7 x-2)^{4}}{28}+C$$

Step-by-Step Solution

Answer

The integral formula is verified correctly by differentiation.

Step 1: Understand the Problem

The exercise requires us to verify the given integral formula by differentiating the right-hand side with respect to \(x\). This means we need to check if the derivative of \(\frac{(7x-2)^4}{28}+C\) is equal to \((7x-2)^3\).

Step 2: Differentiate the Right-Hand Side

Differentiate \(y = \frac{(7x-2)^4}{28} + C\) with respect to \(x\). Use the chain rule for differentiation: \(\frac{d}{dx}[(u(x))^n] = n(u(x))^{n-1} \cdot u'(x)\), where \(u(x) = 7x - 2\), \(n = 4\), and \(u'(x) = 7\).

Step 3: Apply the Chain Rule

Using the chain rule: \[ \frac{d}{dx}\left(\frac{(7x-2)^4}{28}\right) = \frac{1}{28} \cdot 4(7x-2)^{3} \cdot 7 \]Simplify the expression: \[ \frac{1}{28} \cdot 28(7x-2)^{3} = (7x-2)^3 \]

Step 4: Verify the Result

The derivative \((7x-2)^3\) matches the original integrand, confirming that the given integral formula is correct. \(\frac{d}{dx}C = 0\), which means the constant of integration does not affect the verification.

Key Concepts

Chain Rule in Calculus Differentiation

In calculus, the chain rule is a fundamental differentiation technique used when dealing with composite functions. A composite function is where one function is nested inside another.

In simpler terms, if you have a function inside another function, the chain rule helps you find the derivative. To apply the chain rule, follow these steps:

• Identify the outer function and the inner function. In our example,
  • the outer function is \((u(x))^n = (7x - 2)^4\) (raised to the fourth power)
  • the inner function is \(u(x) = 7x - 2\).
• The derivative of the outer function is \(n(u(x))^{n-1}\), which is \(4(7x-2)^3\).
• Multiply the result by the derivative of the inner function, \(u'(x) = 7\).

The chain rule allows you to differentiate complex expressions systematically and is crucial for verifying integrals and handling real-world applications in calculus.

Integral Verification in Calculus

Integral verification is the process of checking if a given antiderivative correctly reverses the differentiation process to yield the original function or integrand.
The goal of integral verification is to ensure the accuracy of calculus operations by confirming the inter-relationships between derivatives and integrals.

In the exercise, we are tasked to verify if the function \(\frac{(7x-2)^4}{28}+C\) when differentiated, returns the original integrand \((7x-2)^3\).

By successfully verifying the integral through differentiation, we confirm the formula's correctness and are assured that it correctly represents the area under the curve described by the original function.This creates a powerful check that provides a mathematical 'double-check' of our integral computation.

Understanding Derivatives in Calculus

Derivatives in calculus are a measure of how a function changes as its input changes. They're at the heart of calculus because they describe how quantities change.
This concept is essential in the world of mathematics, physics, engineering, and other fields.

• A derivative, represented by \(\frac{dy}{dx}\) or \(f'(x)\), is the rate at which \(y\) changes with respect to \(x\).
• It is also the slope of the tangent line to the function at any point.

In our exercise, the derivative of \(\frac{(7x-2)^4}{28}+C\) is calculated using the chain rule. This gives us \((7x-2)^3\), showing how the expression changes with each small change in \(x\).
Understanding derivatives is crucial for graph analysis, calculating speed, optimizations, and more.

Basics of Calculus Integration

Integration in calculus is the process of finding an integral, which represents the accumulation of quantities over a certain interval. It’s often seen as the reverse operation to differentiation.
By integrating a function, you can determine areas under curves, displacement from velocity, and other accumulated changes.

• An integral, \(\int f(x) \, dx\), is the antiderivative of \(f(x)\).
• The process of integration gives us a new function plus a constant, represented by \(C\), because derivatives of constants are zero.

The exercise's given solution \(\int (7x-2)^3 \, dx = \frac{(7x-2)^4}{28} + C\) is an example of finding an antiderivative.

Integral calculus is widely used in science and engineering for problems involving motion, areas, volumes, and more. Mastering integration techniques and their verification is key to understanding the full scope of calculus.