Problem 13 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(13-16,\) verify that \(\int f(u) d u \neq \int f(u) d x\) $$f(u)=\sqrt{u}\( and \)u=x^{2}(x>0)$$

Step-by-Step Solution

Answer

\(\int f(u) du\) and \(\int f(u) dx\) are not equal. Verified.

Step 1: Calculation of \(\int f(u) du\)

The function f(u) is given as \(\sqrt{u}\). So by applying the power rule for integration (Given function \(u^{n}\), its integral is \((u^{n+1})/(n+1) + C)\), we have \(\int u^{1/2} du = (2/3) u^{3/2} + C\). This is the integral of f(u) respect to u

Step 2: Calculation of \(\int f(u) dx\)

Firstly, we replace u by \(x^{2}\) in f(u), to get \(f(x^{2}) = \sqrt{x^{2}} = x\). Now, the integral of x in respect to x is \(\int x dx = (1/2) x^{2} + C\).

Step 3: Comparison of results

Looking at the results from steps 1 and 2, we have \(2/3 u^{3/2} + C\) and \((1/2) x^{2} + C\). These are not equal, thus the equality does not hold, verifying that \(\int f(u) du \neq \int f(u) dx\).

Key Concepts

Integration Techniques

Mastering the art of integration, a fundamental component of calculus, involves familiarity with a variety of methods. These techniques streamline the process of integrating complex functions, a task that often cannot be approached through straightforward application of a basic formula.

• The power rule is commonly used when the function can be expressed as a variable raised to a power.
• U-substitution, a form of variable substitution, is applied when an integral contains a composite function or when it simplifies the integral into a more recognizable form.
• Integration by parts is another method and it follows the product rule for differentiation.
• Techniques like partial fraction decomposition are used to integrate rational functions.
• For trigonometric integrals, strategies involve using trigonometric identities.

Applying these techniques correctly can significantly simplify the task of finding the antiderivative of a function.

Power Rule for Integration

The power rule is a fundamental theorem in calculus, providing a quick formula for finding the integral of power functions. According to the power rule, to integrate a function of the form \( u^n \), where 'n' is a real number, and 'u' is a differentiable function of 'x', the antiderivative is given by:

\[ \frac{u^{n+1}}{n+1} + C \]
where C represents the constant of integration. This rule cannot be applied when 'n' is equal to -1, as this leads to a logarithmic function instead. In the exercise, we used the power rule to integrate \( u^{1/2} \), yielding \( (2/3)u^{3/2} + C \), thus demonstrating the practical application of this technique.

Indefinite Integrals

An indefinite integral, illustrated with the notation \( \int f(x) dx \), represents the antiderivative of a function, and is essential in calculus for finding functions from their rates of change or gradients. What distinguishes indefinite integrals from definite integrals is the absence of boundaries that limit the area calculation.

The result of an indefinite integral includes a constant term 'C', symbolizing the set of all possible antiderivatives. As seen in the power rule example, after integrating \( u^{1/2} \) the result was expressed as \( (2/3)u^{3/2} + C \), where 'C' could represent any real number. The concept of indefinite integrals is vital in solving problems related to accumulation and the reverse process of differentiation.

Variable Substitution

When faced with an integral that cannot be easily solved with standard techniques, variable substitution, commonly referred to as 'u-substitution', can be a powerful tool. This technique involves replacing a portion of the integral with a new variable 'u', which simplifies the integral, making it easier to evaluate.

The process includes choosing 'u', differentiating it to find 'du', and substituting these into the integral. If done correctly, the resulting integral in terms of 'u' should be easier to solve. Once the integration is complete, reverting back to the original variable completes the process. As illustrated in the given exercise, variable substitution clarifies why \( \int f(u) du \) is not equivalent to \( \int f(u) dx \), as each represents a different relationship between the variables.