Definite integrals (part 5) More examples of using definite integrals to calculate the area between curves
Definite integrals (part 5)
- Welcome back.
- Where I had just left off we were trying to figure out this
- area between these two curves, and we figured out that it's
- really the area between the curves between the point 0,
- x equals 0 and x equals 1.
- And I was proposing of a way to do it.
- Let's figure out the entire area under the square root of
- x from 0 to 1, and we can subtract from that, this
- purple area, which is the area under x squared.
- So just based on the last example we did, we could just
- write the indefinite integral, and I'm not going to rewrite
- the fundamental theorem from calculus, because I think
- you know that by now.
- Let me do it in a loud color.
- So I want to know the larger area, right.
- The area under just the square root of x.
- That's Kind of like the combined area.
- Well that's just from 0 to 1, the integral of square root of
- x, because square root of x is the green function, dx.
- And I want to subtract from that the area from 0 to 1
- what's under x squared.
- x squared dx.
- And I just want to make a point.
- This could have just been rewritten this way.
- You could just rewrite a new function, which is the
- difference of these two functions, and it would
- have been equivalent.
- You could have said this.
- This isn't kind of a step of the problem, but you could
- have done it this way.
- In fact, some people start this way.
- See those are the same thing as the integral from 0 to 1
- of square root of x minus x squared dx.
- So you could do this two separate problems, two separate
- indefinite integrals, or you could do it as one
- indefinite integral.
- Actually that might be even simpler because when you
- evaluate it from 1 to 0 it simplify things a little bit.
- So let's stick with the second one.
- So first of all we just have to figure out what the
- antiderivative of this inner expression is.
- So you haven't seen square root of x yet.
- Do you think you know how to do it?
- Well, I think you do.
- Let's say that equals square root of x is just x to
- the 1/2 power, right?
- So we just use the same antiderivative rules
- we've always used.
- We raise it one more power so it becomes x to the 3/2.
- Right, it was 1/2, we added 1 to it.
- And then we divide by this new exponent.
- So dividing by a fraction is like multiplying
- by its reciprocal.
- So it's 2/3 x to the 3/2 and then minus-- I think the second
- term is pretty easy for you-- minus 1/3 x to the third.
- That's the antiderivative of minus x squared, minus
- 1/3 x to the third.
- And we're going to have to evaluate this thing at 0 and 1
- and subtract the difference.
- Subtract this expression evaluated at 0 from this
- expression evaluated at 1.
- I think I'm running out of space.
- What happens when x equals 1?
- 1 to the 3/2 is 1.
- 1 to the third is 1.
- So it's 2/3 minus 1/3.
- Well that's easy.
- It's 1/3.
- I just put 1 in for x.
- And then when x is equal to 0 what is this expression equal?
- Well that's easy too.
- That's 0.
- So there you go.
- 1/3 minus 0 or 1/3.
- That's kind of neat.
- You know I find is this of exciting because if just my
- intuition I was like, oh I have these two curves, and I mean
- they do intersect at the nice integer number, but you know
- what it's probably going to be some really messy number
- of what the areas between these two curves, right.
- Who knows, maybe it'll involve some you know-- a circle
- involves pi, which is this really messy number, so maybe
- all curves have these kind of messy areas.
- But this one it's just one of those neat things about math.
- The area between the square root of x and x squared is 1/3,
- which is a pretty clean number.
- Actually let me do one more problem since I have time.
- It's a bit of a trick problem.
- I mean, you might actually find this easy, but let's figure out
- the area between f of x is equal to-- I don't
- know, x to the fifth.
- I'm going to do something simple.
- Let me draw it actually.
- OK, I'm going to draw the x-axis.
- x to the fifth is going to go up super fast,
- something like that.
- It's going to go up real fast.
- Let's say I wanted to figure out the areas-- this side's
- going to go real fast too-- between that, and instead of
- figuring out the area between x-axis and that, f of x, I want
- to figure out the area between f of x and-- Instead
- of figuring out this bottom area, right?
- Like the normal problems we've done, we would figure out this
- type of area, you know, between two points.
- Let's say I want to figure out the area inside of the curve
- where the height here is 32.
- So I want to figure out this area inside the curve.
- How do we do that?
- Well one way we could do it is just like we did the last one,
- we can figure out some function that's essentially a line, a
- horizontal line that goes straight across here.
- And we'll essentially just be figuring out the area
- between the two functions.
- So what's a function that's a line that just goes
- straight at y equals 32?
- I think I just gave you the answer.
- Let me stick with that greenish color.
- So we could say g of x is equal to 32.
- It's just a constant function.
- It just goes straight across.
- And then we need to figure out what the area is between the
- two, so we need to figure out what are these two points.
- So when does x to the fifth equal 32?
- I mean you could solve it algebraically, you know,
- you could say x to the fifth equals 32.
- x is equal to 2, and actually, you know what?
- I made a mistake.
- Let's say that this is not equal to x to the fifth.
- Let's say that f of x is equal to x to the absolute
- value of x to the fifth.
- Because the mistake, obviously x to the fifth is not a
- parabola looking thing. x to the fifth goes
- negative like this.
- But I have committed so much to this cup shape that I'll make
- it the absolute value of x to the fifth.
- So if I say the absolute value of x to the fifth is equal to
- 32, I think you see where I realized my mistake.
- But if I say the absolute value of x to the fifth is 32,
- there's two places where that's true.
- It's x is equal to plus or minus 2.
- These are the two points.
- I should have done something with an even exponent so I
- could have had this cup shape, but anyway the absolute
- value solved my problem.
- So what is this area?
- We know it's between negative 2 and 2, so we just set up
- the indefinite integral.
- It's the indefinite integral for minus 2 to 2 of the top
- function, the top boundary, 32, minus the bottom boundary.
- Well, this will be a little bit tricky, but minus the absolute
- value of x to the fifth dx.
- And actually instead of doing this, I think you could see
- that there's symmetry here, so we could just figure out
- this area and multiply by 2.
- This problem's a little hairy, just because I had a bad
- choice of initial function.
- Not exactly what I wanted, but we'll work on forward.
- So instead of doing that, let's do the integral from 0 to 2 of
- 32 minus x to the fifth dx.
- And then multiply that by 2.
- So what is that?
- That's 32x minus x to the sixth over 6.
- And we're going to evaluate it from 2 and 0 at 64 minus 64/6.
- 2 to the sixth is 64, and then 32 times 0 is 0 and the
- next is 6, that's 0, so the answer is 64 minus 64/6.
- I'm about to run out of time.
- That's just a fraction problem there.
- Oh, and that's half of it, right?
- So we want to multiply that by 2.
- So if we multiply that by 2, we get 128 minus 128/6.
- I haven't figured out what it is.
- Well I guess we could figure it out.
- It's 128 times 1 minus 1/6 or 128 times 5/6.
- And I don't know what that is.
- I can multiply if I wanted to, but I have 10 seconds left
- so I'll leave you there.
- Hope I didn't confuse you.
- See you soon.
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