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. $$\int_{-1}^{2}(|x|-1)^{3} d x$$

Short answer

The value of the integral \(\int_{-1}^{2} (|x|-1)^{3} dx\) is 4.

Step-by-step solution

Split the integral according to the absolute function

The integral is split into two at \(x = 0\), rewriting \(|x|\) as \(-x\) when \(x < 0\) and \(x\) when \(x \geq 0\). So the given integral \(\int_{-1}^{2} (|x|-1)^{3} dx\) becomes \(\int_{-1}^{0}(-x -1)^3 dx + \int_{0}^{2}(x - 1)^3 dx\)

Expand the cubic term

The cubic term in both integrals is expanded using the binomial theorem to simplify the integral. The expansions are \((-x -1)^3 = -x^3 - 3x^2 - 3x - 1\) and \((x - 1)^3 = x^3 - 3x^2 + 3x - 1\)

Evaluate the integrals

Now, evaluate each integral term individually by applying the power rule for integration which states that the integral of \(x^n\) is \(1/(n+1) * x^{n+1}\). The result is as follows: \[ \int_{-1}^{0}(-x^3 - 3x^2 - 3x - 1) dx = [-\frac{1}{4}x^4 - x^3 - \frac{3}{2}x^2 - x]_{-1}^{0} \] and \[ \int_{0}^{2}(x^3 - 3x^2 + 3x - 1) dx = [\frac{1}{4}x^4 - x^3 + \frac{3}{2}x^2 - x]_{0}^{2} \]

Calculate the definite integrals

The calculated functions are now evaluated at the upper and lower limits of each integral. The result of the first integral is \(2-(-2) = 4\) and the result of the second integral is \((4 - 8 + 6 - 2) - (0) = 0\)

Final Step: Sum the results

Finally, the value of the original integral is the sum of the results from the two calculated integral terms, which is \(4 + 0 = 4\)