WEBVTT
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Okay, So for this problem or given an integral
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equation, uh, Y of X equals four plus
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integral from zero to ex, uh, to tee
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times a squared of y f t d t.
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And at first glance, this might see kind of
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intimidating, I guess, Uh, since you have
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X and T and inter girls and all that.
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But actually, we can make this really simple using
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some fundamental theorem of calculus. So there's fundamental rules
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that calculus always follows. Um, so you start
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with the first thing we do is we take the
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derivative of both sides. So this would become do
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you the derivative y of X? With respect to
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X, you can just write as a dy DX
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equals. So during this would be 00 plus,
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um, and then you have the derivative with respect
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to X of this entire integral to D squared of
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Iot G. And basically, this is the important
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part here because the first fundamental theorem of calculus first
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rule basically says that the derivative of any integral right
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from a to B or eight x of a function
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, it's basically going to be equal to this dysfunction
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here. So what that's also telling us is that
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the derivative f prime of X to differentiate this side
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? If I integrate this side, then I'll get
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obviously capital F, which is one step higher.
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But the derivative would just be the normal thing.
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So from what we can tell from this is we
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can take this and we can apply that directly to
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here. And you can say that dy dx it's
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going to be equal to two X. So the
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derivative of this is just equal to the function and
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the function hit here on the inside is to t
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, uh, squared of y of t So we
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just plug X in there and we'll get to x
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times the square root of y of X uh,
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D X right. We don't need the d.
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X because the derivative and here will cancel out.
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So you'll have just two t times the square root
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of y of X, uh, two x Times
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Square to buy back to my back. Um,
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so now what this is telling us is, uh
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y of X. We can just rewrite as just
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why? Because in this case, that's all it
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is. So you have two x times word of
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y And now we can rewrite this, separating them
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right, applying the separation rules. So we take
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the dy DX and separate them so you can divide
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this side by squared of y and multiply this side
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by DX And what do that you'll get dy over
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the square root of y equals two x dx.
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And, uh, now, finally, the last
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thing I have to do is simply integrate both sides
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. So you integrate this side, uh, with
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respect to the y and you integrate this side with
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respect to the X Uh so another way you can
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be right. This is square, uh, integral
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of why two negative one half dy equals and an
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integral of two x is really, really simple.
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So you can just go ahead and write X squared
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. Plus c don't forget that constant of integration.
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Um, so yeah, so you have X squared
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. Plus, C uh, you have actually received
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there and then on this side for the why it's
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a negative one half. Uh, you just simply
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do the normal power rules, right? X two
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negative, one half plus one and then on the
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bottom, you're going to put negative one half plus
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one as well. Uh, so that will basically
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give you equals X squared plus c and then the
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plus See the integral integral in the constant of integration
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you get from this integral, uh, basically just
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add onto here. So in reality, you have
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, like, a C one and C two here
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, but it's all just a constant. So C
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one plus C two is implied in this plot.
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One single plus c over here. So you don't
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need to worry about both constant of integration. You
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can just combine them. Uh, and after all
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this, you'll end up with basically this will go
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to the top two times y to the one half
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mhm equals X squared plus c. And why did
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the one half is just a fancy way of saying
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square root of why? So you can go out
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and say that. And now if we separate out
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why, then we can simply say, uh,
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why it would be equal to X squared plus c
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over two. And then this whole thing squared Mhm
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. Okay, but now we can actually substitute X
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equals zero to our initial function here, which was
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this so if you say X equals zero, then
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because we need to find this plus see here.
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So we say X equals zero. So are we
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. Just rewrite real quick. Our original, uh
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, function here it was to t times the square
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of y f T d t. And if we
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plug in Y equals zero, then you'll get four
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plus interval from 0 to 0. That's really key
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. Uh, of two tee Times Square, the
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Via T d. T. Now, if you
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remember the rule of integration where if you're going from
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0 to 0, then your integral, which is
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basically zero. So why have zero, which is
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equal to four? So now that we know why
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zero is equal to four, uh, we can
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use this. We just found out why here.
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So we can plug y of zero is equal to
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four, which is equal to zero squared, which
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is zero. And then you'll see over to see
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over two square. Uh, so we take the
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square root of both sides of this little thing here
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, and you're left with two equals C over two
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. And that means C is equal to four.
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So now that you know that you can finally go
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in and plug your C back into your final equation
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, which will change colors here for, uh,
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red. Why is equal to mhm? Was it
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X squared? Plus X squared. Plus 4/2 squared
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. So that would be your final answer. Uh
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, for evaluating this integral.