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- [Instructor] So we're told that f(x)
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is equal to the infinite series,
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we're going from n equals one to infinity
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of n plus one over four to the
n plus one, times x to the n.
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And what we wanna figure out is, what is
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the definite integral from
zero to one of this f(x)?
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And like always, if you feel inspired,
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and I encourage you to feel inspired,
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pause the video and see if
you can work through this
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on your own, or at any time
while I'm working through it,
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pause it and try to keep on going.
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Well, let's just rewrite
this a little bit.
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This is going to be the same thing
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as the integral from zero to one.
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F(x) is this series, so I could write
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the sum from n equals one to infinity
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of n plus one over four to the n plus one,
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times x to the n.
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And now what I'm about
to do might be the thing
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that might be new to some of you,
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but this is essentially, we're taking
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a definite integral of a sum of terms.
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That's the same thing as taking the sum
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of a bunch of definite integrals.
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Let me make that clear.
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So if I had a, let's say
this is a definite integral
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zero to one, and let's say
I had a bunch of terms here.
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I could even call them functions.
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Let's say it was g(x) plus h(x),
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and I just kept going
on and on and on, dx,
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well, this is the same thing
as a sum of the integrals,
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as the integral from zero to one of g(x),
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g(x) dx plus the integral
from zero to one h(x) dx,
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plus, and we go on and on and on forever.
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However many of these terms are.
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This comes straight out of
our integration properties.
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We can do the exact same thing here,
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although we'll just do it
with the sigma notation.
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This is going to be equal to the sum
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from n equals one to infinity
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of the definite integral
of each of these terms.
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So I'm gonna write it like this.
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So of the integral from zero to one
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of n plus one over four
to the n plus oneth power,
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times x to the n, and then it is dx.
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Once again, now we're taking
the sum of each of these terms.
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Let's evaluate this
business right over here.
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That is going to, I'll
just keep writing it out.
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This is going to be equal to the sum
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from n equals one to
infinity, and then the stuff
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that I just underlined in orange,
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this is going to be, let's see,
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we take the antiderivative here.
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We are going to get to
x to the n plus one,
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and then we're gonna divide by n plus one.
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So we have this original n plus one over
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four to the n plus one,
and that's just a constant
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when we think in terms of x, for any one
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of these terms, and then
here we'd wanna increment
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the exponent, and then divide
by that incremented exponent.
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This just comes out of, I
often call it the inverse,
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or the anti-power rule, or
reversing the power rule.
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So it's x to the n plus
one over n plus one.
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I just took the antiderivative,
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and we're gonna go from zero
to one for each of these terms.
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Before we do that, we can simplify.
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We have an n plus one,
we have an n plus one,
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and so we can rewrite all of this.
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This is going to be the same thing,
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we're gonna take the sum from
n equals one to infinity,
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and this is going to be,
what we have in here,
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when x is equal to one, it is one,
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we could write one to the n plus one
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over four to the n plus one.
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Actually, yeah, why don't
I write it that way.
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One to the n plus one over
four to the n plus one,
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minus zero to the n plus one
over four to the n plus one,
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so we're not gonna even
have to write that.
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I could write zero to the n plus one
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over four to the n plus one,
but this is clearly just zero.
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And then this, and this is starting
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to get nice and simple now, this is going
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to be the same thing, this is equal
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to the sum from n equals one to infinity.
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And we almost are gonna
get to our drumroll
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of 1/4 to the n plus one.
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Now you might immediately recognize this.
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This is an infinite geometric series.
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What is the first term here?
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Well, the first term is, well,
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when n is equal to one,
the first term here
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is 1/4 to the second power.
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Did I do that right?
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Yeah.
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When n is equal to one, it's going to be,
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so this is going to be
1/4 to the second power,
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which is equal to 1/16,
so that's our first term.
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And then our common ratio
here, well that's gonna be,
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well, we're just gonna
keep multiplying by 1/4,
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so our common ratio here is 1/4.
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And so for an infinite geometric series,
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this is, since our common ratio,
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its absolute value is less than one,
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we know that this is going to converge,
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and it's gonna converge to the value,
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our first term, 1/16, divided by
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one minus the common ratio, one minus 1/4,
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so this is 3/4, so it's
equal to 1/16 times 4/3.
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This is going to be equal to 1/12.
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And we're done.
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And this seemed really daunting at first,
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but we just have to realize, okay,
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an integral of a sum,
even an infinite sum,
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well, that's gonna be the sum
of these infinite integrals.
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We take the antiderivative
of these infinite integrals,
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which we were able to do, which is kind
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of a cool thing, one of the
powers of symbolic mathematics,
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and then we realized, oh, we just have
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an infinite geometric series,
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which we know how to find the sum of.
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And we're done.