Question

Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = –x^3 and y = –x.

Answer 1

and your work?

Answer 2

i only know how to solve for single functions

Answer 3

I disagree. You're technically subtracting 0 (which is itself a function, albeit a constant one) from any given function value to get the height of a rectangle. The same principle works here, but instead of \(f(x^*)-0\), you have \(f(x^*)-g(x^*)\).

Answer 4

Technicality aside:

Notice that due to symmetry, the area of the bounded region in the second quadrant is the same as the area of the region in the fourth quadrant. So,
\[\text{total area}=\text{area in Q2}+\text{area in Q4}=2\times\text{area in Q2 (or Q4)}\]
or, in terms of integrals,
\[\text{area}=\int_{-1}^0(-x-(-x^3))\,dx+\int_0^1(-x^3-(-x))\,dx=2\int_0^1(x-x^3)\,dx\]