Question

The rate of depreciation of a building is given by D'(t) = 4000(10 − t) dollars per year, 0 ≤ t ≤ 10; see the following figure.

A coordinate plane is given. The horizontal axis is labeled tand the vertical axis is labeled D'(t). A line and a shaded region are graphed.

• The line labeled D'(t) starts at (0, 40000), goes down and right, passes through (5, 20000), and stops at t = 10 on the positive t-axis.
• The area under the line from t = 0 to t = 5 is shaded.

(a) Use the graph to find the total depreciation of the building over the first 5 years

(t = 0

to

t = 5).

$  

(b) Use the definite integral to find the total depreciation over the first 5 years.
$

Answer 1

SOLUTION:-

(a) The area of the shaded portion from t=0 to t=5 is, area of the large triangle - small triangle

The area of the large triangle from t=0 to t=10 is, (1/2) x base x height = (1/2) x 10 x 40000 = 200000 square unit.

  ( Area of a Righ triangle is, (1/2)XBASE X HEIGHT)

The area of the small traingle i.e, from t=5 to t=10 is, (1/2) x base x height = (1/2) x 5 x 20000 = 50000 unit.  (Here base = 10 - 5 =5, and height is 20000)

So area of the shaded portion is, 200000-50000=150000 unit.

Hence the total depreciation of the building over the first 5 years is, $ 150000

(b) Using definite integral, the area formula is,

$\int_{t=0}^{t=5}D(t)dt=\int_{t=0}^{t=5}4000(10-t)dt=\int_{t=0}^{t=5}(40000-4000t)dt$

  $=[40000t-4000\left (\frac{t^{2}}{2} \right )]_{t=0}^{t=5}$

$=[40000(5-0)-4000\left (\frac{5^{2}}{2}-\frac{0^{2}}{2} \right )]$

  

25 = = [40000(5) - 4000 0 2

$=200000-100000/2=200000-50000=150000$

Hence again the total depreciation of the building over the first 5 years is, $ 150000