Sometimes a seemingly easy problem can get fairly complicated with the addition of a few extra requirements. For instance, suppose we want to find the area between the functions *y* = *x*^{2} – 4 and *y* = 3*x*. If we graph the two functions, we see that they appear to cross at *x* = -1 and *x* = 4.

We can verify these two points by setting the functions equal to each other and solving for x.

The area between these curves lies above the parabola and below the line.

To find the area of the shaded region, take the definite integral from *x* = -1 to *x* = 4 of the higher function minus the lower function,

We can evaluate this integrand using the Fundamental Theorem of Calculus,

Let’s now modify this problem by finding the area enclosed by the functions *y* = *x*^{2} – 4 and *y* = 3*x* as well as the lines *x* = -5 and *x* = 1. Graph each of these equations.

The region enclosed by these graphs is more complicated since the functions cross at *x* = -1.

To find the area of the enclosed region, we need to break it into two parts. The first part runs from *x* = -5 to the point of intersection at *x* = -1. The area of this part is

The second part extends from *x* = -1 to *x* = 1 and has area

The area enclosed by the functions *y* = *x*^{2} – 4 and *y* = 3*x* as well as the lines *x* = -5 and *x* = 1 is the sum of these parts or ^{206}/_{3}. Adding the vertical lines on either side of the point of intersection requires the use of two definite integrals since the parabola is higher on the left side of *x* = -1 and the line is higher on the right side of x = -1.

Thus the area between the curves is ^{184}/_{3} +^{ 22}/_{3} or ^{206}/_{3}.