Find the average value of the function on the interval [ 4 , 9 ] \left\lbrack4,9\right\rbrack [ 4 , 9 ] . F ( x ) = x + 6 F\left(x\right)=\sqrt{x}+6 F ( x ) = x <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"> + 6

this problem we're asked to find the average value of the function on the interval from four to nine, where this is the function we're given. Now this is an average value problem since that's in the word here, and if we want to find the average value, well we need this equation. The average value of the function will be one over b minus a multiplied by the integral from a to b of our function f of x integrated with respect to x. If we can solve using this equation and we're familiar with it, then we can solve any type of these average value problems. Now first off, I'm gonna deal with this b minus a. This b minus a refers to our interval. This being the b, the high bound, and then a being the low bound. So we're gonna have one over, in this case, nine minus four. And it's gonna be multiplied by the integral from four to nine. We go from low to high of our function f of x, which is this function up here. So we have the square root of x plus six, and all that's gonna be integrated with respect to x. Now at this point I can simplify things. We have one over nine minus four, which is the same thing as one over five, and it's gonna be multiplied by the integral from four to nine of this function we have right here square root x plus six. Now at this point, what I can do is apply the fundamental theorem of calculus part two, which says that I can integrate this whole thing by taking the antiderivative first and then applying the bounds. Now see we have these two pieces being added together, so I'm gonna do is integrate these separately. Now we'll start with the integral of the square root of x. So I have one fifth, and that's gonna be multiplied by the the antiderivative of the square root of x. And if you go through the process of taking this antiderivative here, you need to change this to x to the one half power, and then do the power rules here. But I'm just gonna show you the result. This antiderivative is gonna come out to two times x to the three halves power over three. This is what happens when you take the antiderivative of the square root of x. Then we'll have plus the antiderivative of six, which is six x. And all of this is gonna be bounded from four to nine. Now at this point we need to plug in our bounds. So So we'll start by plugging in nine. Now I can see for this first portion, I'm going to have one fifth that's always gonna be out front here, then that's gonna be multiplied by two times nine to the three halves power, all divided by three plus six times nine. That's gonna be this first portion here. Then what I'm gonna do is subtract off. What happens when we plug in the low bound? So we plug in the high bound, and then we'll plug in the low bound of four. So we're gonna have one fifth on the outside still, and then it's gonna be two times, and then we'll plug in four. So we'll have four to the three halves power, all divided by three plus, and then we'll have six times four. And that's what happens when we plug in our high and low bound respectively. Now what you can do is crunch these numbers. You can do this on a calculator. You can try to do this by hand because you actually will get a fraction when you do this. The whole thing should come out to a 28 over 15 as a fraction. So one twenty eight over 15 would be our solution here. And on a calculator, this is the same thing as approximately 8.533 repeating. So we'll just write that as 8.53. So this would be our result. So this is the solution to the problem. That's how you can find this integral from the average value of the function equation. So that's how you can solve these types of problems where you're looking for the average value. Hope you found this video helpful.

8. Definite Integrals

Average Value of a Function

Business Calculus

8. Definite Integrals

Average Value of a Function

Struggling with Business Calculus?

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Multiple Choice

Find the average value of the function on the interval $\left\lbrack4,9\right\rbrack$ .
$F\left(x\right)=\sqrt{x}+6$

11.26

4.74

10.67

8.53

Verified step by step guidance

Step 1: Recall the formula for the average value of a function on an interval [a, b]. It is given by: $ \text{Average Value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx $.

Step 2: Identify the given function $ f(x) = \sqrt{x} + 6 $ and the interval $[4, 9]$. Here, $ a = 4 $ and $ b = 9 $.

Step 3: Substitute the values of $ a $, $ b $, and $ f(x) $ into the formula: $ \text{Average Value} = \frac{1}{9-4} \int_{4}^{9} (\sqrt{x} + 6) \, dx $.

Step 4: Break the integral into two parts for easier computation: $ \int_{4}^{9} (\sqrt{x} + 6) \, dx = \int_{4}^{9} \sqrt{x} \, dx + \int_{4}^{9} 6 \, dx $.

Step 5: Compute each integral separately: $ \int_{4}^{9} \sqrt{x} \, dx $ involves using the power rule for integration, and $ \int_{4}^{9} 6 \, dx $ is a constant multiple. After finding these values, substitute them back into the formula to calculate the average value.