Find the area of the region in the xy-plane bounded above by the graph of the function f(x)=(1/2)x−1, below by the x-axis, on the left by the line x=2, and on the right by the line x=16. The area is

Question

Find the area of the region in the xy-plane bounded above by the graph of the function f(x)=(1/2)x−1, below by the x-axis, on the left by the line x=2, and on the right by the line x=16. The area is ?

Answer 1

Greetings Mindy! Let us solve this problem: Area under the Curve!

the function is labeled as the upper bound of the area and y = 0 is labeled as the lower bound when we draw horizontal line y = 0 (same as drawing x-axis). The Area must equal the height times the width. The height is upper bound minus the lower bound such that [(1/2)x-1] - [0].

We then set the area up to an integral of the Area as

A = (integral from 2 to 16) [(1/2)x-1] dx, where dx is the width. We evaluate the integral with the anti-derivative to obtain,

(integral from 2 to 16) [(1/2)x-1] dx = (x²/4) - x evaluated at 2 to 16

= [(16)²/4 - (16)] - [(2)²/4 - (2)]

= 256/4 - 16 - 1 + 2

= 64 - 16 - 1 + 2

= 48 - 1 + 2

Area = 47 + 2 = 49 square units