Question

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

Short answer

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

Step-by-step solution

- 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 \(F(t)\).

- 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.

- 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\[f'(x) = 1\]

- 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 \(f(x) = x\)