Definite Integrals (part 4) Examples of using definite integrals to find the area under a curve
Definite Integrals (part 4)
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- Welcome back.
- I'm now going to use definite integrals to figure out the
- areas under a bunch of curves and, if we have time, maybe
- even between some curves.
- So let me right down the fundamental theorem
- of calculus.
- I know I covered it really fast in the last presentation.
- Just to make sure you understand this formula.
- The last couple presentations were really to give you an
- intuition for this exact formula.
- Let's say that big f prime of x is equal to f of x, right?
- That's also like saying that the -- that's equivalent to
- saying that f of x -- big f of x -- is equal to the
- antiderivative of f of x, right?
- Well, let's just that it's one of the possible antiderivatives
- of f of x, right?
- Because there's always a constant term here and you're
- not sure whether it is.
- And this is why people tend to use this standard because we
- know that f of x is the derivative of big f prime of x.
- Big f of x is just one of the antiderivatives of f of x.
- So this is a little bit not true, but I think
- you get the idea.
- But the fundamental theorem of calculus tells us if this top
- line is true, then the definite integral from a to b of f of x
- d x is equal to its antiderivative evaluated at b
- minus its antiderivative evaluated at a.
- And I know I said here that big f isn't the only
- antiderivative, right?
- Because you could at any constant to this and that would
- also be the antiderivative.
- But when you subtract here, the constants will cancel out.
- So it really doesn't matter which of the
- constants you pick.
- The constant actually doesn't matter.
- So that's why I actually said the antiderivative.
- But let's apply this.
- You might be confused right now.
- So let me draw a graph.
- There you go.
- Look how straight that is.
- Draw the x-axis.
- Not perfect but it'll do.
- Let's say that my f of x is equal to x squared plus 1.
- So f of x looks like this.
- This is 1.
- So it'll start at 1.
- It'll just be a parabola.
- Let me see how good I can draw this.
- I've done worse.
- So that's f of x.
- It's a parabola. y-intercept at 1.
- And let's say I want to figure out the area under the curve --
- between the curve and, really, the x-axis.
- Let's say I want to figure out the area between the curve and
- the x-axis from x equals negative 1 to, I don't
- know, x equals 3.
- So this is the area I want to figure out.
- I'm going to shade it in.
- So this is the area.
- All of this stuff.
- I want to figure out this area.
- And you could imagine, before you knew calculus, figuring out
- an area of something with a curve -- it's kind
- of top boundary.
- It would have been very difficult.
- But we will now use the fundamental theorem of calculus
- and hopefully you have an intuition of why this works and
- how the integral is really just a sum of a bunch of little,
- little, small squares with infinitely small bases.
- But if you watched the last videos, hopefully that
- hit the point home.
- But now we'll just mechanically compute because, actually,
- understanding it is a bit harder than just doing it.
- But let's just mechanically compute it.
- So we are essentially just going to figure out the
- integral from minus 1 to 3 of f of x, which is x
- squared plus 1 d x.
- What's the antiderivative of x squared plus 1?
- This just equals the antiderivative.
- So it's just x to the third -- we could say 1/3 x to the third
- or x to the third over 3 -- plus x, right?
- The derivative of x is 1.
- And then we don't have to worry about plus c because we're
- going to subtract out the c's.
- You'll see.
- I think you'll get the point.
- It doesn't matter.
- You could pick an arbitrary c right here and it'll
- just cancel out.
- And we're going to evaluate that at 3 and negative 1 and
- we're going to subtract out big f of negative
- 1 from big f of 3.
- This is just the notation they use.
- You figure out the antiderivative and
- you say where you're going to evaluate it.
- And then this is equal to -- So if I evaluate 3.
- 3 to the third power is what?
- That's 27.
- 27 divided by 3 is 9.
- And then 9 plus 3 is 12.
- This is just big f of 3, right?
- Because I figured out the end -- This is big f of x.
- You can kind of view this as big f of x.
- But not to be confused with small, cursive f of x.
- This is big f of x.
- So this big f of 3.
- And then, from that we'll subtract big f
- of negative 1, right?
- Minus big f of negative 1.
- And if we put minus 1 here.
- Let's see, minus 1 to the third power is minus 1.
- So it's minus 1/3 and then plus minus 1, right?
- So minus 1/3 plus minus 1.
- I think that equals minus 4/3, correct?
- I think so.
- Maybe I'm making a mistake with negative signs.
- Minus 1/3 minus 4/3.
- And I'm going to subtract that, right?
- So if I'm subtracting minus 4/3, it's the same thing
- as adding minus 4/3.
- And then we have our answer.
- Actually, it's 12 and 4/3 -- whatever -- units.
- Squared units.
- 12 and 4/3 squared units.
- We could write this as a mixed number as well.
- Let's do another one.
- I'll do a slight variation.
- Let me draw again.
- Some coordinates.
- I don't know if I'm going to have time to do it in
- this video but I'll try.
- I always try.
- Let's say I have f of x is equal to the square root of x.
- So it looks something like this.
- That's actually a pretty nice looking, kind of sideways
- parabola, I think.
- This is f of x.
- And let's say I have another function.
- g of x which equals x squared.
- So g of x is actually going to look something like this.
- I was doing well and then something happened.
- And, of course, it'll continue on this side as well.
- Because it is defined for negative numbers.
- But anyway, my question to you, or my question to myself,
- really, is what is the area between the curves
- where they intersect?
- What is this?
- What is this area?
- Well, the first thing you have to figure out is just what
- are the boundary points?
- What is this point?
- And what is this point?
- Well this point, I think, is pretty clear.
- It's 0, 0, right?
- They both intersect at 0, 0.
- And even this point, you could probably do it from intuition.
- But if you don't, I guess, want to do it through intuition, you
- could just set these 2 equations equal to
- each other, right?
- You could say x squared is equal to the square
- root of x, right?
- And then you could do a bunch of things.
- You could square both sides or -- well, actually, this is the
- same thing as doing it by intuition.
- But I think it's pretty obvious that the only places where x
- squared is equal to the square root of x are the points x
- equals 0, which you already know, and x equals 1.
- So this is the point 1, 1.
- Which is true for both of them.
- And this is more algebra, so I won't go into that
- in too much detail.
- I'm kind of running out of time.
- So we want to figure out the area between these 2 curves.
- So what we can do is -- maybe you want to pause it and think
- about it yourself -- we can figure out the area
- under the grey curve.
- We could figure out this area.
- So we want to figure out -- this is a boundary, right?
- Between 0 and 1.
- We could figure out this area and then we could figure out
- the entire area under the green curve separately and then we
- could subtract the difference.
- Which is exactly how we're going to do it in the next
- video because I have run out of time.
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