lecture33 - 6.5 Applications of the Definite Integral...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: § 6.5 Applications of the Definite Integral Math 121 Lecture 33 ! Average Value of a Function Over an Interval ! Consumers’ Surplus (*) ! Future Value of an Income Stream ! Volume of a Solid of Revolution Determine the average value of f ( x ) = 1 – x over the interval -1 ! x ! 1. Using (2) with a = -1 and b = 1, the average value of f ( x ) = 1 – x over the interval -1 ! x ! 1 is equal to An antiderivative of 1 – x is . Therefore, So, the average value of f ( x ) = 1 – x over the interval -1 ! x ! 1 is 1. ( Average Temperature ) During a certain 12-hour period the temperature at time t (measured in hours from the start of the period) was degrees. What was the average temperature during that period? The average temperature during the 12-hour period from t = 0 to t = 12 is ( Average Temperature ) During a certain 12-hour period the temperature at time t (measured in hours from the start of the period) was degrees....
View Full Document

{[ snackBarMessage ]}

Page1 / 5

lecture33 - 6.5 Applications of the Definite Integral...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online