Question

Evaluate. Then interpret the result in terms of the area above and/or below the X-axis. 1 / 1 f (x3 - 2x) dx 2 2 17 5 (43 - 2x) dx = (Type an integer or a simplified fraction.)

Answer 1

The integral can be decomposed and written as

$\int ^{\frac{1}{2}}_{-\frac{1}{2}}(x^3-2x)dx=\int ^{\frac{1}{2}}_{-\frac{1}{2}}x^3dx-2\int ^{\frac{1}{2}}_{-\frac{1}{2}}xdx$

Now,

$\int ^{\frac{1}{2}}_{-\frac{1}{2}}x^3dx=\frac{x^4}{4}|^{\frac{1}{2}}_{-\frac{1}{2}}=\frac{(\frac{1}{2})^4}{4}-\frac{(-\frac{1}{2})^4}{4}$$=0$

And

$\int ^{\frac{1}{2}}_{-\frac{1}{2}}xdx=\frac{x^2}{2}|^{\frac{1}{2}}_{-\frac{1}{2}}=\frac{(\frac{1}{2})^2}{2}-\frac{(-\frac{1}{2})^2}{2}=0$

Hence,

$\begin{gathered} \int ^{\frac{1}{2}}_{-\frac{1}{2}}(x^3-2x)dx=\int ^{\frac{1}{2}}_{-\frac{1}{2}}x^3dx-2\int ^{\frac{1}{2}}_{-\frac{1}{2}}x=0-2(0) \\ \end{gathered}$$\boxed{\therefore\int ^{\frac{1}{2}}_{-\frac{1}{2}}(x^3-2x)dx=0}$