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## Integral Calculus

### Unit 3: Lesson 4

Area: vertical area between curves- Area between a curve and the x-axis
- Area between a curve and the x-axis: negative area
- Area between a curve and the x-axis
- Area between curves
- Worked example: area between curves
- Area between two curves given end points
- Area between two curves
- Composite area between curves

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# Worked example: area between curves

AP.CALC:

CHA‑5 (EU)

, CHA‑5.A (LO)

, CHA‑5.A.1 (EK)

The area between the graphs of functions ƒ and 𝑔 can be found by taking the definite integral of ƒ-𝑔 (suppose ƒ is above 𝑔).

## Want to join the conversation?

- At5:18, why is the area positive if most of the graph is below the x-axis?(13 votes)
- Area is always positive, the negative values of the integrals make it positive although the if you're finding only the value of integral, the negative is accepted.(5 votes)

- After1:30he says that when finding area b/w curves....it makes sense to subtract one equation from another and then integrate. He also said that there are many videos in which he has discussed that why it makes sense. Could someone pls tell me which videos is he talking about(5 votes)
- Could you take the anti-derivative of the two functions in the beginning and then subtract them?(4 votes)
- How can we find the lower and upper bounds without referring to the graph?(3 votes)
- If you have equations solved for the same variable, you can set them equal to each other in order to find the points on intersections. These points are your bounds.(3 votes)

- I got the same answer taking the difference between the definite integral of the "upper" function and the "lower" function (definite integral of X^4 +4x^2+1 minus definite integral of x^2-3)

I don't understand why the first definite integral give me a negative value (precisely, -4/15). I expected a positive value, representing the area between the x axys and the blue function(3 votes) - Is it possible to find the area between two curves
*without*looking at their graphs?(2 votes)- If you can solve for where the functions intersect each other and determine which function is greater than the other in each interval between intersections, then you would not need to look at their graphs before integrating.(3 votes)

- how can you tell top vs. bottom(3 votes)
- The one that is above the other over the interval is the top.(0 votes)

- Could someone please explain what to do if you have two equations, y=2lnx and y=x-3, and how to get the coordinate points needed?(1 vote)
- Generally, an equation with one side linear and the other side logarithmic cannot be solved algebraically for x. So it's best to use a graphing calculator or equation solver to find the x-coordinate(s) of the intersections of the graphs of y=2lnx and y=x-3.(4 votes)

- Can somebody help me, please? I'm trying to solve the below equation for x, and I'm on a step that's giving me confusion. When I see step 1, it's clear that one of the solutions for x is -7, but that solution applies no longer when I move on to step 2. What's happening??

Step 1: (x+7) = 0.25(x+7)^2

step 2: 1 = 0.25(x+7) ## divided both sides by the same amount, (x+7).(1 vote)- The reason your equation is breaking is that you're dividing by 0. x+7=0 when x=-7.(3 votes)

- How can you tell top and bottom without looking at the graph?(1 vote)
- Simply plug values into each of the functions. Whichever function has a larger output must be on top.(3 votes)

## Video transcript

- [Instructor] What we're going to do using our powers of calculus is find the area of this yellow region and if at any point you get inspired, I always encourage you to pause the video and try to work through it on your own. So, the key here is you
might recognize hey, this is an area between curves. A definite integral might be useful, so I'll just set up the
definite integral sign and so, first we're gonna
think about what are left and right boundaries of our region? Well, it looks like the left boundary is where the two graphs
intersect right over here and the right boundary is where they intersect right over there. Well, what is this point of intersection? It looks like it is
negative one, negative two. We can verify that. In this red curve, if X is negative one, let's see, you square that,
you'll get one minus three. You do indeed get Y is
equal to negative two and in this blue function was X is equal to negative one, you get one minus four plus one. Once again, you do indeed get
Y is equal to negative two and the same thing is true when X is equal to one. One minus three, negative two, one minus four plus one, negative two. So, our bounds are
indeed, we're going from X equals negative one to
X equals positive one and now let's think about
our upper and lower bounds. Over that interval this blue graph is our upper bound and so, we would subtract the lower bound from the upper bound, so we would have X to the fourth, minus four X squared plus one and from that we will subtract X squared minus three DX and in many other videos we have talked about why you do this, why this makes sense to just subtract the lower graph from the upper graph when you're finding the area between them but now we just have to
evaluate this definite integral, so let's just get down to business. Alright, so we have the integral
from negative one to one, and so, we have X to the
fourth, X to the fourth, and now we have minus four X squared and then when you distribute
this negative sign, you're gonna subtract another X squared, so you're gonna have minus five X squared and then you have plus one and then you're gonna
subtract a negative three, so it's gonna be one plus three, so it's gonna be plus four DX, DX and just to be clear I should put parentheses right over there because it's really the
DX is being multiplied by this entire expression and so, let's see, let's find
the antiderivative of this. This should be pretty straightforward. We're just gonna use
the reverse power rule multiple times, so this is
going to be the antiderivative of X to the fourth is X
to the fifth over five. We just incremented the exponent and divided by that incremented exponent, minus, same idea here, five X to the third over three plus four X and then we are going to evaluate it at one and then subtract from that and evaluate it at negative one. So, let's first evaluate it at one. We're gonna get 1/5 minus 5/3 plus four and now let us evaluate
it at negative one, so minus, let's see, if
this is negative one, we get a negative 1/5, negative 1/5 and this is gonna be plus 5/3, plus 5/3 and then this is
going to be minus four, minus four but then when you distribute the negative sign, we're
going to distribute this over all of these terms, and so this is going to be, if we make this positive,
this will be positive, this one will be negative and then this one will be positive, so you have 1/5 plus 1/5 which is going to be 2/5, that and that, and then minus 5/3, minus 5/3, so minus 10 over three and then four plus four, so plus eight and so, we just need to simplify this. This is going to be, let's see, it's going to be eight and then if I write, so plus, I'm gonna write these two
with a denominator of 15 'cause that's the common
denominator of three and five. Let's see, 2/5 is 16/15,
yeah, that's right, five times three is 15. Two times three is six and then 10/3, let's see, if we multiply the denominator times fives, we have to multiply the
numerator times five, so this is going to be 50/15 and so, what's 6/15 minus 50/15? So, this is going to be equal to eight minus six minus 50 is minus 44, minus 44 over 15 and so what is 44 over 15? 44 over 15 is equal to two and 14/15, so that's really what we're subtracting. We're gonna subtract two and 14/15, so if you subtract two from this you would get six minus 14 over 15 'cause we still have to subtract the 14/15 and then six minus 14/15 is going to be equal to five and 1/15. So, just like that, we were indeed able to figure out this area.