Consider the region bounded above by f(x) =x, below by the x-axis, on the left by x= 0,and on the right by x= 5. The region has been divided into five vertical strips.

Question

Write the definite integral representing the exact area, and evaluate that integral

Answer 1

desmos.com/calculator/bctqfppeg7

The above Desmos graph allows you to enter a function, endpoints of an interval [a, b], number of sub-intervals (n), and to choose left, right, or midpoint rule.

Answer 2

You can work this problem with geometry and trig, or integral calculus

it doesn't hurt to do both to check your answer

with geometry&trig:

exact area = the area of a right triangle with base =5, height = 5

area = hb/2 = 5(5)/2 = 25/2 = 12.5 = 12 1/2 square units

with calculus:

integral of xdx = (x^2)/2

evaluate from 0 to 5

= 5^2/2 -0^2/2 = 25/2 = 12.5= 12 1/2 units^2

by geometry:

each of 5 vertical strips would have base =1

height at the middle of each strip would = 1/2, 3/2, 5/2, 7/2 and 9/2

area of each strip would also = 1/2, 3/2, 5/2, 7/2 and 9/2

sum = (1+3+5+7+9)/2 = 25/2 = 12 1/2 un.^2

or by calculus:

integral of xdx for each strip = (x^2)/2

evaluate for 1st strip from 0 to 1, for 2nd from 1 to 2, for 3rd from 2 to 3, for 4th from 3 to 4, for 5th from 4 to 5

= (1^2 -0^2 + 2^2 -1^2 + 3^2 - 2^2 + 4^2 -3^2 + 5^2 -4^2)/2

=(1 + 3 + 5+ 7 + 9)/2

= 25/2

= 12 1/2 un^2

f(x)=x is the same as y=x, a 45 degree line out of the origin

it helps to graph the line y=x and visually look at the area bounded by it and the x axis and the vertical line x=5. they enclose a right triangle. the line x=0 is the y axis and really doesn't add to the restrictions already given ; it's "surplusage"