Problem 44 - Essential Calculus

Question

Find a Maclaurin series for \(f(x)\). $$ f(x)=\int_{0}^{x} \sqrt{1+t^{3}} d t $$

Step-by-Step Solution

Answer

The Maclaurin series for the function \(f(x) = \int_{0}^{x} \sqrt{1+t^{3}} dt \) is \(f(x) = x\)

Step 1 - Find the Indefinite Integral

To start, we need to find an anti-derivative \(F(t)\) of the function inside the integral \(\sqrt{1+t^{3}}\). However, there's no elementary function whose derivative is \(\sqrt{1+t^{3}}\). Hence, we cannot explicitly find \(F(t)\).

Step 2 - Differentiating then Integrating

Instead of directly finding the anti-derivative, we could differentiate under the integral sign per the Leibniz rule. From the Leibniz rule, we can write that,\[f'(x) = \sqrt{1+x^{3}}\]This gives the derivative of the function \(f(x)\). Now, by integrating the derivative, we get the original function \(f(x)\), because the integral of a derivative is the original function.

Step 3 - Maclaurin Series of the Derivative

The Maclaurin series for \(f'(x)\) is an expansion of \(f'(x)\) as an infinite sum of terms calculated from the values of its derivatives at a single point. The general form of a Maclaurin series for a function is\[f'(x) = f'(0) + f''(0) x + \frac{f'''(0)}{2!} x^{2} + \frac{f''''(0)}{3!} x^{3} + \ldots\]Given our function, \(f'(0) = \sqrt{1+0^{3}} = 1\)This also implies \(f(0) = 0\)Any derivative of \(\sqrt{1+x^{3}}\) will be in the form \(\frac{p(x)}{(1+x^{3})^{q}}\), where \(p(x)\) is a polynomial. Hence, if we substitute \(x=0\), we get 0. Therefore, \(f''(x) = 0, f'''(x) = 0\), and so on, and the general series will be of the form\[f'(x) = 1\]

Step 4 - Maclaurin Series of the Original Function

Now, we know the Maclaurin series of derivative \(f'(x)\). To find the Maclaurin series for \(f(x)\), we integrate the series for \(f'(x)\). The indefinite integral of \(f'(x)\) is just \(f(x)\), which is \(x\). Hence the Maclaurin series for \(f(x)\) is \(f(x) = x\)

Key Concepts

Indefinite Integral

Finding an indefinite integral involves determining the antiderivative of a function. This process essentially reverses differentiation, allowing us to recover a function from its rate of change. However, not all functions have easy-to-find antiderivatives expressed by elementary functions.
In the case given, \[ f(x)= \int_{0}^{x} \sqrt{1+t^{3}} dt \]the expression \(\sqrt{1+t^{3}}\) doesn't have a straightforward antiderivative. When basic techniques do not yield a solution, alternate strategies must be employed, like the Leibniz Rule or numerical methods. This is a key point to understand when studying indefinite integrals: not every integral is easily solvable with familiar functions.

• Indefinite integrals bring functions back from their derivatives.
• They help find "original" functions without boundary conditions.
• Sometimes, alternative methods are needed when standard ones are not viable.

Leibniz Rule

The Leibniz Rule is a powerful tool for handling integrals, especially when dealing with parameters of integration that depend on variables. It provides a method for differentiating integrals with respect to a parameter. In simple terms, by using this rule, we can differentiate under the integral sign to find derivatives of integrals with variable limits.
For the problem at hand, the derivative of the integral was found using:\[f'(x) = \sqrt{1+x^{3}}\]which was achieved by recognizing that the upper limit of the integral, \(x\), acted as a variable. Hence, applying the Leibniz Rule simplified finding the derivative directly from the integral function without needing the antiderivative of \(\sqrt{1+t^{3}}\).

• The rule works well when standard integration techniques fail.
• It's valuable for solving integrals with variable limits.
• Facilitates the computation of derivatives indirectly linked to integrals.

Derivatives

Derivatives are fundamental in calculus, representing the rate of change of a function concerning its variable. They form the backbone of various concepts, including the Maclaurin series, which helps to express a function as an infinite sum.
In the scenario detailed, the derivative of the function was expressed as \(f'(x) = \sqrt{1 + x^3}\). To find the Maclaurin series, the approach involved calculating derivatives at the point \(x = 0\). For this function: - \(f'(0) = 1\),- further derivatives give zero, such that these subsequent terms don't contribute to the series.

• Derivatives describe how functions change, allowing for deeper analysis.
• They are crucial for formulating Taylor and Maclaurin series.
• In scenarios like this, integration and differentiation work hand-in-hand to solve complex problems.

Understanding derivatives is essential in mastering calculus applications like series expansion and solving differential equations.