Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 7

Applications of the Definite Integral

Net Change

In this type of problem, we must identify a rate of change. The key to doing so is to pay attention to the units.

Example 7.4. [See Exercise 58 on page 318 of the textbook.]

Water flows from the bottom of a storage tank at a rate of ${\displaystyle r\left(t\right)=200-4t}$ litres per minute, where ${\displaystyle 0\le t\le 50\text{.}}$ Find the amount of water that flows from the tank during the first 10 minutes.

The given rate is ${\displaystyle r\left(t\right)=200-4t}$ litres/min, and its antiderivative is the amount of water ${\displaystyle w\left(t\right)}$ at time ${\displaystyle t\text{.}}$ The amount of water during the first 10 min is ${\displaystyle w\left(10\right)-w\left(0\right)}$; hence,

${\displaystyle w\left(10\right)-w\left(0\right)={\int }_0^{10}200-4tdt=200t-2t^2{|}_0^{10}=1800.}$

The amount of water is ${\displaystyle 1800}$ litres.