Question

Use the value \(\int_{0}^{2} x^{3} d x=4\) to evaluate each definite integral. Explain your reasoning. (a) \(\int_{-2}^{0} x^{3} d x\) (b) \(\int_{-2}^{2} x^{3} d x\) (c) \(\int_{0}^{2} 3 x^{3} d x\)

Short answer

(a) \(\int_{-2}^{0} x^{3} dx = -4\), (b) \(\int_{-2}^{2} x^{3} dx = 0\), (c) \(\int_{0}^{2} 3x^{3} dx = 12\)

Step-by-step solution

Evaluate \(\int_{-2}^{0} x^{3} dx\)

The integral \(\int_{-2}^{0} x^{3} dx\) is the negative of \(\int_{0}^{2} x^{3} dx\) due to the symmetry of \(x^3\). Thus, if \(\int_{0}^{2} x^{3} dx = 4\), it follows that \(\int_{-2}^{0} x^{3} dx = -4\).

Evaluate \(\int_{-2}^{2} x^{3} dx\)

The integral \(\int_{-2}^{2} x^{3} dx\) can be broken down into two integrals as \(\int_{-2}^{2} x^{3} dx = \int_{-2}^{0} x^{3} dx + \int_{0}^{2} x^{3} dx\), applying the property of symmetry of the odd function we get this integral to be \(0\).

Evaluate \(\int_{0}^{2} 3x^{3} dx\)

The integral \(\int_{0}^{2} 3x^{3} dx\) is three times \(\int_{0}^{2} x^{3} dx\) due to the constant multiple rule of integrals. Since \(\int_{0}^{2} x^{3} dx = 4\), it follows that \(\int_{0}^{2} 3x^{3} dx = 3*4 = 12\).