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. $$\int_{-1}^{1} x^{3} d x$$

Short answer

The value of the integral \(\int_{-1}^{1} x^3 dx\) is 0.

Step-by-step solution

Determine the nature of the function

Check if \(x^{3}\) is an odd or even function. A function is odd if \(f(-x) = -f(x)\). Here if we replace \(x\) in \(x^{3}\) with \(-x\) we get \((-x)^3 = -x^3\), which means it is an odd function.

Understand the implications of an Odd Function

For an odd function \(f(x)\), the integral \(\int_{-a}^{a} f(x) dx = 0\), as the area under the graph from -a to 0 will be negative (below x-axis) and from 0 to a will be positive (above x-axis), and they will cancel out each other when we add them together.

Evaluate the Integral

Applying the result from Step 2 to the given integral, \(\int_{-1}^{1} x^3 dx = 0\).