Find the area of the region in the xy-plane

Question

Find the area of the region in the xy-plane bounded above by the graph of the function f(x)=(4/11)x−(12/11), below by the x-axis, on the left by the line x=3, and on the right by the line x=14.

 

The area is     ?    

Answer 1

This question is essentially asking you to find the area under the curve f(x) = (4/11)x - 12/11 between x = 3 and x = 14. To do that, take the integral of the function on the interval [3, 14]:

 

∫[3, 14] ( (4/11)x - 12/11 ) dx = (2/11)x² - (12/11)x | [3, 14] = (2/11)(196) - (12/11)(14) - (2/11)(9) + (12/11)(3) = 392/11 - 168/11 - 18/11 + 36/11 = 242/11 = 22 square units