Average Values of Functions on Intervals

In summary, the conversation discusses two problems involving finding the average values of functions. The first problem involves finding the average value of y for the curve y = 3x-x^2 in the first quadrant, with the answer being -6. The second problem is about finding the average value of cos x on the interval [-3,5], with the answer being (sin3 +sin5)/8. The conversation also includes a brief discussion on evaluating antiderivatives and the use of identities in simplifying expressions.
  • #1
Tiome_nguyen
5
0

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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  • #2
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 . ^^
 

FAQ: Average Values of Functions on Intervals

What is the "problem with average value"?

The "problem with average value" refers to the limitations and potential inaccuracies of using the mean, or average, to represent a set of data. It is important to understand that the average is just one measure of central tendency and may not accurately represent the entire dataset.

What causes the problem with average value?

The problem with average value can be caused by outliers, or extreme values, in the dataset. These outliers can skew the average and make it an inaccurate representation of the majority of the data. It can also be caused by a non-normal distribution of the data, where the majority of the values are not clustered around the mean.

How can the problem with average value be addressed?

One way to address the problem with average value is to use alternative measures of central tendency, such as the median or mode. These measures are less affected by outliers and can provide a more accurate representation of the data. Additionally, it is important to visually inspect the data and understand its distribution before relying solely on the average.

Why is it important to be aware of the problem with average value?

Being aware of the problem with average value is important because it can lead to incorrect conclusions and decisions based on the data. For example, if outliers are not identified and removed, the average may appear to be higher or lower than it actually is, leading to inaccurate interpretations of the data.

How does the problem with average value impact scientific research?

The problem with average value can have a significant impact on scientific research. If the average is used as the sole measure of central tendency, it can lead to incorrect conclusions and potentially invalidate the research findings. It is important for scientists to thoroughly understand and appropriately address the limitations of the average when analyzing and interpreting data.

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