Integral of Two Triangles

Hckyplayer8

Junior Member
Joined
Jun 9, 2019
Messages
205
1.PNG

I sketched the function which is a diagonal line. I placed [-2,2] as my bounds and found f(-2) = -3 and f(2) = 5. They served as the height of my two triangles.

My next task was to find the x intercept of the function as that designated the change from negative area to positive area. I found the x intercept to be -1/2. Then it was just a matter of finding the x coord distance between -2 and -1/2, as well as -1/2 and 2 to serve as the base of the triangles.

I labeled the triangle with positive area as A1 and the triangle with negative area as A2.

Thus for A1 = [2.5 * 5] / 2 = 6.25 and for A2 [3*1.5] / 2 = 2.25

Then for the final answer A1 - A2 = 6.25 - 2.25 = 4

Does this look reasonable?
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
5,295
Yes, everything looks just right.
 

Hckyplayer8

Junior Member
Joined
Jun 9, 2019
Messages
205
Sweet! Appreciate it.
 
Top