Question

Use \(\int_{0}^{2} x^{2} d x=\frac{8}{3}\) to evaluate the definite integrals without using the Fundamental Theorem of Calculus. (a) \(\int_{-2}^{0} x^{2} d x\) (b) \(\int_{-2}^{2} x^{2} d x\) (c) \(\int_{0}^{2}-x^{2} d x\) (d) \(\int_{-2}^{0} 3 x^{2} d x\)

Short answer

(a) -8/3, (b) 0, (c) -8/3, (d) -8.

Step-by-step solution

Evaluate integral (a)

Simply changing the limits of integration from [0,2] to [-2,0] will change the sign of the integral. Thus, \(\int_{-2}^{0} x^{2} dx = -\int_{0}^{2} x^{2} dx = - \frac{8}{3}\)

Evaluate integral (b)

The integral over [-2,2] can be split into two separate integrals, namely \(\int_{-2}^{0} x^{2} dx + \int_{0}^{2} x^{2} dx\). From step 1 we know that \(\int_{-2}^{0} x^{2} dx = - \frac{8}{3}\) and we are given \(\int_{0}^{2} x^{2} dx = \frac{8}{3}\). Therefore \(\int_{-2}^{2} x^{2} dx = - \frac{8}{3} + \frac{8}{3} = 0\)

Evaluate integral (c)

The given integral \(\int_{0}^{2}-x^{2} dx\) is equal to \(- \int_{0}^{2} x^{2} dx\), meaning the integral expression is multiplied by -1. Therefore, \(\int_{0}^{2}-x^{2} dx = - \int_{0}^{2} x^{2} dx = - \frac{8}{3}\)

Evaluate integral (d)

The integral \(\int_{-2}^{0} 3x^{2} dx\) can be seen as \(3 \cdot \int_{-2}^{0} x^{2} dx\), meaning we can take the 3 outside the integral and multiply it with \(\int_{-2}^{0} x^{2} dx = -\frac{8}{3}\) Therefore, \(\int_{-2}^{0} 3x^{2} dx = 3 \cdot -\frac{8}{3} = -8\)