Question

Use the value \(\int_{0}^{1} x^{2} d x=\frac{1}{3}\) to evaluate each definite integral. Explain your reasoning. (a) \(\int_{-1}^{0} x^{2} d x\) (b) \(\int_{-1}^{1} x^{2} d x\) (c) \(\int_{0}^{1}-x^{2} d x\)

Short answer

(a) \(\int_{-1}^{0} x^{2} d x = \frac{1}{3}\) \n (b) \(\int_{-1}^{1} x^{2} d x = \frac{2}{3}\) \n (c) \(\int_{0}^{1}-x^{2} d x = -\frac{1}{3}\)

Step-by-step solution

Calculation of \(\int_{-1}^{0} x^{2} d x\)

The integrand is \(x^2\), and it's an even function. This means the integral from -a to a is twice the integral from 0 to a. Since the exercise only asks for the integral from -1 to 0, it would be the same as the integral from 0 to 1, so \(\int_{-1}^{0} x^{2} d x = \frac{1}{3}\).

Calculation of \(\int_{-1}^{1} x^{2} d x\)

As stated in Step 1, the integrand \(x^2\) is an even function, so the integral from -1 to 1 is twice the integral from 0 to 1. So, \(\int_{-1}^{1} x^{2} d x = 2 \times \frac{1}{3} = \frac{2}{3}\).

Calculation of \(\int_{0}^{1}-x^{2} d x\)

Here the integral is of \(-x^2\) from 0 to 1. We know the integral of \(x^2\) from 0 to 1 is \(\frac{1}{3}\). Then the integral of \(-x^2\) from 0 to 1 would be \(-1 \times \frac{1}{3} = -\frac{1}{3}\).