Find the average value of the function on the interval [ 0 , 1 ] \left\lbrack0,1\right\rbrack . G ( x ) = 2 x 2 + 1 G\left(x\right)=\frac{2}{x^2+1}

in this problem, we're asked to find the average value of the function on the interval from zero to one, where this is the function that we're given. So let's see how we can solve this problem. Now when it comes to finding the average value of functions, there's an equation that we can use. The average value is gonna be one over b minus a multiplied by the integral from a to b of our function f of x with respect to x. Now notice in this problem the function we're dealing with is g of x. So I'm actually going to take this and make it g of x instead. It really ultimately doesn't matter too much from the standpoint that we just know to plug in our function into this equation, but this is what's gonna be a little more accurate for the problem that we're solving. Now to go ahead and calculate the average value of this function, well, first off, we're gonna take a look at this interval. This one here would refer to the b, the high term, then the zero would refer to the a. So going to right here we'd have one over b minus a, which is the same thing as one over one minus zero, multiplied by the integral from zero to one of our function, which in this case is two over x squared plus one, and that's integrated with respect to x. So now that we have this we'll have one over one minus zero which is all just gonna come out to one multiplied by this integral, so we don't even need to rewrite that. So all we're really dealing with is this integral right here. So we're gonna go ahead and use this as what we're solving for. Now let's go ahead and see how we can simplify things. Well, recall that I can use the fundamental rules when it comes to antiderivatives and integrals. And the fundamental theorem of calculus allows me to actually take the antiderivative of this first and then evaluate this at the bounds. So we'll first find the antiderivative, and I'll start by taking this two that I see this constant and pulling it outside of the integral. That's a rule that we've learned about. So we have two multiplied by the integral from zero to one of one over x squared plus one with respect to x. Now how do I go about finding the integral of one over x squared plus one? Well, right at first, this looks quite tricky, but you may recall some of the common derivatives and integrals we learned about in earlier chapters. And one of the things we learned about was the derivative of the inverse tangent of x. And the inverse tangent of x derivative came out to be one over x squared plus one. And if this comes out to be the derivative, remember that taking the antiderivative basically just allows us to go backwards. So this whole thing would technically just be the inverse tangent of x if we integrated it. So if we go ahead and do this, we're going to have two that's gonna be multiplied by the inverse tangent of x based on the rules that we've learned about, and then this whole thing is going to be bounded from zero to one, the balance that we have right there. So that's how we can write this in a more simple way. Now let's go ahead and see what we can do from here. Well, what I'm gonna try doing is applying the balance. We'll start with one. So we're gonna have that's gonna be two on the outside multiplied by the inverse tangent of what we plug in for x, which is one. So I'll plug one in for x, and that's gonna be the first piece here. And then what I need to do is subtract this from our low bound being plugged in. So we'll have two multiplied by the inverse tangent, and now for our low bound, well, we're gonna plug in zero. So I'll put zero in for this inverse tangent, and that's going to be our low bound. So we have two times the inverse tangent of one minus two times the inverse tangent of zero. I know this whole thing is just gonna come out to be zero, and the inverse tangent of one multiplied by two, well that's going to give me pi over two. So pi over two would be our result, and that is the solution to this problem. So this is how you can solve these types of problems where you need to find the the average value of a function, and you're given some kind of tricky situations where we get trig functions that pop out. Even for problems like this, take a look at some of the common derivatives of trig functions that we've seen or inverse trig functions and see if that starts to mimic or look like any of these types of functions that we see as we simplify things. So you apply these tools to integrals, see if something looks familiar, and then you can go ahead and relate that to find the integral, and then you plug in the bounds, and you'll get your results. So I hope you found this video helpful. That's how you can solve this problem. Let's go ahead and move on.

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 $open bracket 0 comma 1 close bracket$ .
$G (x) = \frac{2}{x^{2} + 1}$

$pi over 4$

$π$

$pi over 2$

$0$

Verified step by step guidance

Identify the function for which you need to find the average value. Here, the function is $ G(x) = \frac{2}{x^2 + 1} $.

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

Substitute the given function and interval into the formula: $ \text{Average value} = \frac{1}{1-0} \int_{0}^{1} \frac{2}{x^2 + 1} \, dx $.

Evaluate the integral $ \int_{0}^{1} \frac{2}{x^2 + 1} \, dx $. This integral is a standard form that results in $ 2 \tan^{-1}(x) $ evaluated from 0 to 1.

Substitute the limits of integration into the antiderivative: $ 2 \tan^{-1}(1) - 2 \tan^{-1}(0) $. Simplify this expression to find the average value.