Here, well that's gonna be, well, we're just gonna When n is equal to one, it's going to be, so this is going to beġ/4 to the second power, which is equal to 1/16, The first term here is 1/4 to the second power. What is the first term here? Well, the first term is, well, when n is equal to one, Now you might immediately recognize this. Get to our drumroll of 1/4 to the n plus one. And then this, and this is starting to get nice and simple now, this is going to be the same thing, this is equal to the sum from n equals one to infinity. I could write zero to the n plus one over four to the n plus one,īut this is clearly just zero. Over four to the n plus one, so we're not gonna even One to the n plus one overįour to the n plus one, minus zero to the n plus one What we have in here, when x is equal to one, it is one, we could write one to the n plus one over four to the n plus one. N equals one to infinity, and this is going to be, This is going to be the same thing, we're gonna take the sum from We have an n plus one, and so we can rewrite all of this. I just took the antiderivative, and we're gonna go from zero Often call it the inverse, or the anti-power rule, or Here we'd wanna increment the exponent, and then divideīy that incremented exponent. So we have this original n plus one over four to the n plus one,Īnd that's just a constant when we think in terms of x, for any one of these terms, and then X to the n plus one, and then we're gonna divide by n plus one. Infinity, and then the stuff that I just underlined in orange, this is going to be, let's see, we take the antiderivative here. This is going to be equal to the sum from n equals one to Let's evaluate thisīusiness right over here. To the n plus oneth power, times x to the n, and then it is dx. So of the integral from zero to one of n plus one over four This is going to be equal to the sum from n equals one to infinity of the definite integral We can do the exact same thing here, although we'll just do it On and on and on, dx, well, this is the same thingĪs a sum of the integrals, as the integral from zero to one of g(x), g(x) dx plus the integralįrom zero to one h(x) dx, plus, and we go on and on and on forever. Let's say it was g(x) plus h(x), and I just kept going This is a definite integral zero to one, and let's say That's the same thing as taking the sum of a bunch of definite integrals. To do might be the thing that might be new to some of you, but this is essentially, we're taking a definite integral of a sum of terms. F(x) is this series, so I could write the sum from n equals one to infinity of n plus one over four to the n plus one, times x to the n. This is going to be the same thing as the integral from zero to one. While I'm working through it, pause it and try to keep on going. You can work through this on your own, or at any time Zero to one of this f(x)? And like always, if you feel inspired, and I encourage you to feel inspired, pause the video and see if And what we wanna figure out is, what is the definite integral from So we're told that f(x) is equal to the infinite series, we're going from n equals one to infinity of n plus one over four to the
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