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. $$\int_{0}^{1}\left(x^{3}-1\right) d x$$

Short answer

The value of the integral \(\int_{0}^{1}(x^{3}-1) dx\) is \(-\frac{3}{4}\)

Step-by-step solution

Break Down the Integral

The integral \(\int_{0}^{1}(x^{3}-1) dx\) can be broken down into the sum of two separate integrals. Hence, \(\int_{0}^{1}(x^{3}-1) dx = \int_{0}^{1}x^{3}dx - \int_{0}^{1}dx\).

Applying the Given Fact

The fact given in the exercise – that \(\int_{0}^{1}x^{3}dx = \frac{1}{4} \)– can be applied to the first part of our broken down integral, so we get \(\frac{1}{4} - \int_{0}^{1}dx\).

Solve the Remaining Integral

The integral \(\int_{0}^{1}dx\) is simply the area under the curve from \(0\) to \(1\), which makes it \(1-0=1\). Hence, the result from Step 2 becomes \(\frac{1}{4} - 1\).

Final Calculation

Now, calculate the final result \(\frac{1}{4} - 1 = -\frac{3}{4}\).