Average Values of Functions on Intervals

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SUMMARY

The discussion focuses on calculating the average values of functions over specified intervals. The average value of the curve defined by the equation y = 3x - x² in the first quadrant is incorrectly stated as -6; the correct average value must be positive due to the nature of the parabola. For the average value of cos x over the interval [-3, 5], the correct formula is (sin 5 - sin(-3))/8, which simplifies to (sin 5 + sin 3)/8 using the identity sin(-x) = -sin(x). The participants clarify the correct evaluation of integrals and limits of integration.

PREREQUISITES
  • Understanding of average value of a function
  • Knowledge of definite integrals
  • Familiarity with trigonometric identities
  • Ability to analyze parabolic functions
NEXT STEPS
  • Study the concept of average value of a function in calculus
  • Learn how to evaluate definite integrals, focusing on limits of integration
  • Review trigonometric identities, particularly sin(-x) = -sin(x)
  • Explore properties of parabolic functions and their graphs
USEFUL FOR

Students studying calculus, particularly those focusing on integration and average values of functions, as well as educators seeking to clarify these concepts for their students.

Tiome_nguyen
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Homework Statement



i have some problems that i tried to do but i couldn't get the answer , i hope you can help me, please,

1. what is the average value of y for the part of the curve y = 3x-x^2 which is in the first quadrant? the answer is -6 , but i couldn't get it

2. the average value of cos x on the interval [-3,5] is ? the answer is (sin3 +sin5)/8 , i only got ( sin -3 - sin5)/8 , i have no idea why is sin3 + sin5 .

i hope you can help me with these problem , thank u . ^^



Homework Equations





The Attempt at a Solution

 
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Tiome_nguyen said:

Homework Statement



i have some problems that i tried to do but i couldn't get the answer , i hope you can help me, please,

1. what is the average value of y for the part of the curve y = 3x-x^2 which is in the first quadrant? the answer is -6 , but i couldn't get it
I don't see how the average value could possibly be -6. The graph of y = 3x - x^2 is a parabola that opens downward. Except for the two x-intercepts, the y-values on the portion of the graph in the first quadrant are all positive y-values, so the average value has to be positive. What do you have for your integral?



Tiome_nguyen said:
2. the average value of cos x on the interval [-3,5] is ? the answer is (sin3 +sin5)/8 , i only got ( sin -3 - sin5)/8 , i have no idea why is sin3 + sin5 .
I think you are evaluating the limits of integration for your antiderivative in the wrong order.
You should have gotten (1/8)(sin 5 - sin(-3)). By identity, sin(-x) = -sin(x), for all x.
Tiome_nguyen said:
i hope you can help me with these problem , thank u . ^^
 

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