Problem 68 - Thomas Calculus

Question

Right, or wrong? Give a brief reason why. $$\int \frac{x \cos \left(x^{2}\right)-\sin \left(x^{2}\right)}{x^{2}} d x=\frac{\sin \left(x^{2}\right)}{x}+C$$ Wrong

Step-by-Step Solution

Answer

The given integral solution is incorrect because its derivative does not match the integrand.

Step 1: General Understanding of the Problem

To determine whether the integral \( \int \frac{x \cos \left(x^{2}\right)-\sin \left(x^{2}\right)}{x^{2}} \, dx \) equals \( \frac{\sin \left(x^{2}\right)}{x} + C \), we need to integrate the expression on the left and then compare with the given result.

Step 2: Analyze the Provided Answer Purpose

The given solution to the integral is \( \frac{\sin(x^2)}{x} + C \). To solve the problem, we need to comprehend what happens if we differentiate \( \frac{\sin(x^2)}{x} \) because if its derivative is equal to the integrand, then the provided result is correct.

Step 3: Differentiate the Given Expression

Differentiate \( \frac{\sin(x^2)}{x} \) using the quotient rule, which states that for \( y = \frac{u}{v} \), \( y' = \frac{v\cdot u' - u\cdot v'}{v^2} \). Here, \( u = \sin(x^2) \) and \( v = x \).

Step 4: Find Derivatives of Numerator and Denominator

The derivative of \( u = \sin(x^2) \) is \( u' = \cos(x^2) \cdot 2x \) by the chain rule. The derivative of \( v = x \) is \( v' = 1 \).

Step 5: Apply the Quotient Rule

Using the quotient rule: \[ \frac{d}{dx} \left( \frac{\sin(x^2)}{x} \right) = \frac{x \cdot ( \cos(x^2) \cdot 2x ) - \sin(x^2) \cdot 1}{x^2} = \frac{2x^2 \cos(x^2) - \sin(x^2)}{x^2}. \]

Step 6: Simplify and Compare with Integrand

Simplify the expression \( \frac{2x^2 \cos(x^2) - \sin(x^2)}{x^2} \) to get \( 2x \cos(x^2) - \frac{\sin(x^2)}{x^2} \). This is not equal to the original integrand \( \frac{x \cos(x^2) - \sin(x^2)}{x^2} \).

Step 7: Conclusion

Since differentiating \( \frac{\sin(x^2)}{x} + C \) does not result in the original integrand, the provided integral solution is incorrect.

Key Concepts

Quotient Rule

The Quotient Rule is a fundamental technique in calculus used to differentiate functions that are written as a quotient, that is, a fraction of two functions. If we have a function in the form of \(y = \frac{u}{v}\), where both \(u\) and \(v\) are differentiable functions of \(x\), the derivative \(y'\) is given by:

• \(y' = \frac{v \, u' - u \, v'}{v^2}\)

To apply the quotient rule, we need to calculate the derivatives of the numerator \(u\) and the denominator \(v\) separately.
In our exercise, the numerator \(u = \sin(x^2)\) and the denominator \(v = x\). The derivative of the numerator is found using the chain rule, and the derivative of the denominator is straightforward.
The quotient rule helps us to determine whether the given function shares its integral with another function, by checking if the integration leads back to the function inside the integral.

Chain Rule

The Chain Rule is another differentiation technique you will use when dealing with composite functions. For a function \( f(g(x)) \), where one function is nested inside another, the chain rule states:

• \(\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)\)

In the context of our exercise, the function \( u = \sin(x^2) \) involves using the chain rule. Here, \( g(x) = x^2 \) and \( f(g) = \sin(g) \), thus applying the rule gives us the derivative of the outer function \(\cos(g)\) times the derivative of the inner function, \(2x\).
This rule provides a systematic way to differentiate functions that are built by nesting one function inside another and is crucial in obtaining the derivative in the quotient rule.

Definite Integrals

A definite integral is a way to calculate the area under the curve of a function between two points. It's denoted as \(\int_{a}^{b} f(x) \, dx\). This integral calculates the net area, considering areas above the x-axis as positive and areas below as negative.

• Unlike indefinite integrals, definite integrals result in a specific numerical value.
• The Fundamental Theorem of Calculus connects differentiation and integration, allowing us to evaluate definite integrals using the antiderivative.

Definite integrals are not explicitly involved in the given exercise but knowing them can help understand the difference from indefinite integrals where the answer includes an arbitrary constant \(C\).

Indefinite Integrals

Indefinite integrals represent families of functions and are often used to find the original function given its derivative. They are written as \(\int f(x) \, dx = F(x) + C\).

• The result of an indefinite integral is plus a constant \(C\), reflecting the fact that there are multiple functions with the same derivative.
• This indeterminate form allows flexibility in solving calculus problems where the conditions of the function are not fully specified.

In the exercise, the result \(\frac{\sin(x^2)}{x} + C\) represents such an indefinite integral result, suggesting a family of functions possibly consistent with the antiderivative sought. We used differentiation rules to tell whether the antiderivative is indeed valid, showcasing the need to match the original integrand.