Find the average value of the function
f(x)=8x^2-5x+6 , on the interval [3,5]. Find the value of x-coordinate at which the function assumes it's average value.
what is the average value = to ?
what is the x coordinate = to ?
Thanks
Let f(x)=8x^2-5x+6
The average value is the definite integral divided by the interval.
I=∫f(x)dx
=∫(8x^2-5x+6)dx
=[(8/3)x^3-(5/2)x^2+6x] from 3 to 5
=700/3
Average value, a
= I/(5-3)
= 350/3
To find x where f(x)=a, solve for x in
f(x) = 8x^2-5x+6 = 350/3
However, out of the two solutions, x=-3.42 and x=4.045, you will report the solution which is within the interval [3,5] and reject all solutions outside of the interval.
To find the average value of a function on the interval [a, b], we use the following formula:
Average value = (1 / (b - a)) * ∫[a to b] f(x) dx
First, let's find the average value of the function f(x)=8x^2-5x+6 on the interval [3,5]:
Average value = (1 / (5 - 3)) * ∫[3 to 5] (8x^2-5x+6) dx
Taking the integral of the function, we get:
Average value = (1 / 2) * [ (8/3)x^3 - (5/2)x^2 + 6x ] evaluated from 3 to 5
Simplifying further, we get:
Average value = (1 / 2) * [ (8/3)(5^3) - (5/2)(5^2) + 6(5) ] - (1 / 2) * [ (8/3)(3^3) - (5/2)(3^2) + 6(3) ]
Average value = (1 / 2) * [ (8/3)(125) - (5/2)(25) + 30 ] - (1 / 2) * [ (8/3)(27) - (5/2)(9) + 18 ]
After simplifying the expression, we get:
Average value ≈ 51.5
So, the average value of the function on the interval [3, 5] is approximately 51.5.
Now let's find the value of the x-coordinate at which the function assumes its average value. To do this, we need to set the function equal to the average value and solve for x:
8x^2 - 5x + 6 = 51.5
Rearranging the equation, we get:
8x^2 - 5x - 45.5 = 0
Now we can solve this quadratic equation by factoring or using the quadratic formula:
Using the quadratic formula, we get:
x = (-(-5) ± sqrt((-5)^2 - 4(8)(-45.5))) / 2(8)
Simplifying further, we get:
x = (5 ± sqrt(25 + 1456)) / 16
x = (5 ± sqrt(1481)) / 16
Therefore, the x-coordinate at which the function assumes its average value is given by:
x ≈ (5 + sqrt(1481)) / 16 or x ≈ (5 - sqrt(1481)) / 16
To find the average value of a function on a given interval, you need to take the definite integral of the function over the interval and divide it by the length of the interval.
In this case, the function is f(x) = 8x^2 - 5x + 6, and the interval is [3,5].
First, let's find the definite integral of the function over the interval [3,5]:
∫(from 3 to 5) (8x^2 - 5x + 6) dx
To find this integral, we need to replace every x term with x raised to the appropriate power and divide each term by its new power:
∫(from 3 to 5) (8/3)x^3 - (5/2)x^2 + 6x dx
Now, integrate each term:
(8/3) * (1/4)x^4 - (5/2) * (1/3)x^3 + (6/2)x^2 | (from 3 to 5)
Next, substitute the upper limit (5) into the integral expression and subtract the result of substituting the lower limit (3):
[(8/3) * (1/4)(5^4) - (5/2) * (1/3)(5^3) + (6/2)(5^2)] - [(8/3) * (1/4)(3^4) - (5/2) * (1/3)(3^3) + (6/2)(3^2)]
Simplify this expression to find the value of the definite integral.
Once you have the value of the integral, divide it by the length of the interval [3,5] (which is 5 - 3 = 2) to find the average value of the function.
Finally, to find the x-coordinate at which the function assumes its average value, set up the equation f(x) = average value and solve for x. In this case, set up the equation:
8x^2 - 5x + 6 = average value
Substitute the average value you found earlier into the equation and solve for x. This will give you the x-coordinate at which the function assumes its average value.