Find the average of f(x)=sqrt(1-x) on the interval [-1,1]?

In summary, the formula for finding the average of a function on a given interval is: Average = ∫f(x)dx / (b-a). To find the integral of f(x)=sqrt(1-x), we can use the Power Rule for integration. The lower bound for the given interval is -1, and the upper bound is 1. The average of a function on a given interval can be negative, and there is a shortcut method for finding it known as the Mean Value Theorem for Integrals. This theorem states that the average value of a function f(x) on the interval [a,b] is equal to the function's value at the point c, where c is any value between a and b.
  • #1
Chandasouk
165
0
Please show work. There was a negative sign that appeared in the solution and I don't know why.

I know [tex]\sqrt{}(1-X)[/tex] can be rewritten as (1-X)^1/2 and then you use the power rule to integrate but that gives you

(1-x)^3/2
------------
3/2

2(1-X)^3/2
--------------
3

My book shows

-(1-X)^3/2
------------
3/2

instead
 
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  • #2
You integrate that by using the substitution u=1-x. So dx=(-du). Then use the power rule on the u integral. That's where the '-' sign comes from.
 

FAQ: Find the average of f(x)=sqrt(1-x) on the interval [-1,1]?

What is the formula for finding the average of a function on a given interval?

The formula for finding the average of a function on a given interval is:
Average = ∫f(x)dx / (b-a)
Where f(x) is the function, a is the lower bound of the interval, and b is the upper bound of the interval.

How do you find the integral of f(x)=sqrt(1-x)?

To find the integral of f(x)=sqrt(1-x), we can use the Power Rule for integration, where we raise the exponent by 1 and divide by the new exponent. In this case, the integral would be:
∫sqrt(1-x)dx = (2/3)(1-x)^(3/2) + C

What are the lower and upper bounds for the given interval [-1,1]?

The lower bound for the given interval is -1, and the upper bound is 1.

Can the average of a function on a given interval be negative?

Yes, the average of a function on a given interval can be negative. This can happen if the values of the function are mostly negative within the interval, causing the integral to result in a negative value.

Is there a shortcut method for finding the average of a function on a given interval?

Yes, there is a shortcut method for finding the average of a function on a given interval, called the Mean Value Theorem for Integrals. This theorem states that the average value of a function f(x) on the interval [a,b] is equal to the function's value at the point c, where c is any value between a and b. Therefore, the average value of f(x) on the interval [a,b] can be found by evaluating the function at any point c within the interval.

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