# Problem with average value

## 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 . ^^

## The Attempt at a Solution

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Mark44
Mentor

## 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?

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