In Problems 79-90, use properties of even and odd functions to find each integral. \( \begin{array}{ll}\text { 79. } \int_{-2}^{2}\left(x^{2}-4\right) d x & \text { 80. } \int_{-1}^{1}\left(x^{3}-2 x\right) d x \\ 440 & \text { 82. } \int_{-\pi / 4}^{\pi / 4} \sec ^{2} x d x \\ \text { 83. } \int_{-1}^{1} \frac{3}{1+x^{2}} d x & \text { 84. } \int_{-5}^{5}\left(x^{1 / 3}+x\right) d x \\ \text { 85. } \int_{-\pi / 2}^{\pi / 2} \frac{1}{3} \sin \theta d \theta & |2 x| d x\end{array} \)

To find the integral for problem 79, we can break down the function. The integrand \(x^2 - 4\) is an even function because \(f(-x) = (-x)^2 - 4 = x^2 - 4 = f(x)\). The integral of an even function over a symmetric interval \([-a, a]\) can be computed as twice the integral from \(0\) to \(a\). Thus: \[ \int_{-2}^{2}(x^2 - 4)dx = 2\int_{0}^{2}(x^2 - 4)dx = 2\left[\frac{x^3}{3} - 4x\right]_{0}^{2} = 2\left(\frac{8}{3} - 8\right) = 2\left(\frac{8 - 24}{3}\right) = -\frac{32}{3}. \] Now for problem 80, the integrand \(x^3 - 2x\) is an odd function because \(f(-x) = -x^3 + 2x = -(x^3 - 2x) = -f(x)\). The integral of an odd function over a symmetric interval \([-a, a]\) is always zero. Thus: \[ \int_{-1}^{1}(x^3 - 2x)dx = 0. \] Two integrals down, many more to go!