Find the average value of the function on the interval [ 2 , 5 ] \left\lbrack2,5\right\rbrack . F ( x ) = x 3 − 2 x F\left(x\right)=x^3-\frac{2}{\sqrt{x}}

In this problem, we're asked to find the average value of the function on the interval from two to five, and this is the function we're given. Okay, so how can we go about solving this? Well, my first step here is going to be to write the equation for the average value of a function. And the average value of the function is going to be one over b minus a multiplied by the integral from a to b of our function being integrated with respect to x. Now as you can see we have our function right up here and this is the equation we need, so we have everything we need to solve this problem. Keep in mind b to a refer to the interval, the high and low bounds respectively of the interval that we have. So let's go ahead and plug everything in. So if I wanna find the average value of the function that we have up here, well, first off, we're going to have one over b minus a, which looking at our interval, the high bound is five. The low bound is two. And then that's gonna be multiplied by the integral from two to five, a to b, right there. That's gonna be multiplied by our function f of x. Now you can see that our function right here is x cubed minus two over the square root of x. So we'll have x cubed minus two over the square root of x, and all of this is gonna be integrated with respect to x. Now let's go ahead and see how we can simplify this further. Now first off, I can deal with this fraction which is just one third. I can just simplify this right here. So it's gonna be the start of what we have this, and that's gonna be multiplied by this integral from two to five. And what I can do is use the fundamental theorem of calculus where I go ahead and use take the antiderivative of this function, and then I evaluate the bounds after that. Now I can take the integral separately of each of these pieces of the function. First off, we have this x cubed right here, and the anti derivative of x cubed, well that's gonna give us x to the fourth power over four. Now from here I can subtract off the integral of two over the square root of x. And integrating two over the square root of x, well that's the same thing as integrating two multiplied by x to the one half power. Right? Or actually x to the negative one half power since we have this square root of x in the denominator. And this is gonna be from two to five, but we're just gonna focus on what this antiderivative comes out to be. So we're integrating that with respect to x. And what I can do is pull this two out of the front here. So we're gonna have two times the integral of x to the negative one half power. And then using the power rule, you should get this as two times x to the one half power over one half, which is the same thing as four times the square root of x. So this whole thing, if you integrate it, is going to be four times the square root of x, and then this all going to be bounded from two to five. So this is the result we get so far. Now at this point what we need to do is plug in our bounds and evaluate the integral that we're dealing with. So I can go ahead and go through this process. So first off, we're going to have this one third out in front, and then I'm gonna go ahead and take every place that I see x and replace it with five. So that's going to give us we're gonna have five to the fourth power over four minus and we'll have four times the square root of five. All of this is gonna be part of the first portion here, then we'll subtract off our low bound plug then. And I can see that we're dealing with two, and we have this one third out front again. So we'll have one third. That's gonna be multiplied by and then we'll plug in two for x. So we'll have two to the fourth power over four minus four times the square root of two. And then that's going to be our low bound plug then. So this point, you can evaluate these numbers, and you can do this on a calculator if you want to because the numbers come out a little bit strange. You do get a decimal approximation, which I'm gonna write down here. This should approximately come out to 49.65 as the final result. So this would be the solution to our problem, and that is how we can find the answer for the average value of the function that we had. So hope you found this video helpful, and let's move on and get some more practice.

8. Definite Integrals

Average Value of a Function

Calculus

8. Definite Integrals

Average Value of a Function

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Multiple Choice

Find the average value of the function on the interval $open bracket 2 comma 5 close bracket$ .
$F (x) = x^{3} - \frac{2}{\sqrt{x}}$

29.79

49.65

74.48

107.76

Verified step by step guidance

To find the average value of a function F(x) on the interval [a, b], use the formula: $ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} F(x) \, dx $.

Identify the function and the interval: Here, $ F(x) = x^3 - \frac{2}{\sqrt{x}} $ and the interval is [2, 5].

Set up the integral for the function over the interval [2, 5]: $ \int_{2}^{5} \left( x^3 - \frac{2}{\sqrt{x}} \right) \, dx $.

Calculate the integral: Break it into two parts, $ \int_{2}^{5} x^3 \, dx $ and $ \int_{2}^{5} -\frac{2}{\sqrt{x}} \, dx $, and solve each separately.

After finding the integral, divide the result by the length of the interval, which is $ 5 - 2 = 3 $, to find the average value.

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Struggling with Calculus?

Find the average value of the function on the interval [2,5]\left\lbrack2,5\right\rbrack.F(x)=x3−2xF\left(x\right)=x^3-\frac{2}{\sqrt{x}}

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Average Value of a Function practice set

Find the average value of the function on the interval $open bracket 2 comma 5 close bracket$ .
$F (x) = x^{3} - \frac{2}{\sqrt{x}}$