Question

In Exercises \(47-56,\) use graphs, your knowledge of area, and the fact that \(\quad\) $$\int_{0}^{1} x^{3} d x=\frac{1}{4}$$ to evaluate the integral. $$\int_{0}^{1} \sqrt[3]{x} d x$$

Short answer

Therefore, the value of the integral \(\int_{0}^{1} \sqrt[3]{x} d x = \frac{3}{4}\)

Step-by-step solution

Understand the relationship between the given functions

We need to understand that \(x^{3}\) and \(\sqrt[3]{x}\) are inverse operations of each other. Therefore, the integral of one will inversely correspond to the other. That is, the integral of \(x^{3}\) from \(0\) to \(1\) is \(\frac{1}{4}\), then the integral of \(\sqrt[3]{x}\) from \(0\) to \(1\) will inversely correspond to \(\frac{1}{4}\).

Apply the power rule for integration

The formula for integrating a simple power of \(x\) is \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\), where \(C\) is the constant of integration. But, this case is a bit tricky because we are finding the integral of the cube root of \(x\), which can also be written as \(x^{1/3}\). By applying the power rule, the integral of \(x^{1/3}\) from \(0\) to \(1\) is given by \( \frac{x^{1/3 + 1}}{1/3 + 1} \Biggr|_0^1 \).

Calculate the definite integral

By simplifying, we get \(\frac{x^{4/3}}{4/3} \Biggr|_0^1\). By plugging the limits (0 and 1), we then get the result \(\frac{1^{4/3}}{4/3} - \frac{0^{4/3}}{4/3} = \frac{3}{4}\).