# Area Between Two Curves

Jun 11, 2021
In this video we're going to talk about how to calculate the

## area

between two

#### curves

, so let's go over the basics. Let's say we have some function f of x and we want to find the

## area

under the curve from a to b, then If we are looking for the area of ​​the shaded region, the area is simply the definite integral from a to b of f of x dx now let's say if we have another function from a to b and let's call this function g of x the area under this curve is going to be the integral from a to b but from g of x dx now, what if we want to find the area between two

#### curves

?
So let's say if we have both f of x, let me draw this better and g of x, how can I? we find the area between these two curves, all we have to do is take the difference between this area and this area and we will get the area between f and g of x, so the area is simply the integral from a to b f of x minus g of x dx this is how you can find the area between the two curves you need to take the difference between the top function and subtract it by the bottom function and then take the definite integral of that difference and you will get the area of ​​that curve now let's say if we have another function we'll call this function f of y and let's say it varies from c to d along the y axis and we want to calculate the area of ​​c to d between that curve and the axis and the area of ​​that region will be the integral of c to d of f of y d and then that function is basically x is equal to some function of and now let's say if we want to find the area between two curves then let's say we have f of y and g of y go from c to d, so instead of taking the top function and subtracting by the bottom function, what we're going to do is do is take the integral from c to d of the function on the right, which is f of y minus the function on the left, which is g from y, this is how we can get the area between two curves from c to d when we use values ​​of y, so remember that when it comes to f of x, y is equal to f of x when it comes to f of y your x is equal to f of y now let's go ahead and work on some practice problems let's calculate the area of ​​the region bounded by the line y is equal to 8 minus 2x the x axis and the y axis well, let's start by making a drawing so that y is equal to 8 minus 2x, let's get the 0, this will be 8 minus 2 times 0, which is 8. and If we replace y with 0 and solve for x, we can see that x will be 8 divided by 2, which is 4.

## area between two curves...

So we have a y-intercept of 8 and an x-intercept of 4. So we want to find the area between this line the x-axis and the y-axis so one way to do this is by using geometry we know that the area of ​​a triangle is half the base times the height the base is four the height is eight half of two that is, half of four is two times eight, that's 16. So that's the area of ​​the shaded region, but let's use calculus to get this answer also for that we know that the area between a curve and the x axis will be the integral from a to b of f of x dx and as we know y is equal to f of x and y is eight minus two x, so f of x can be replaced with eight minus two x and along the x axis we are going to integrate it from 0 to 4. so we are going to have the integral of o from 0 to 4 f of of 2x to the first power is 2x squared over 2. so we can cross out the 2. so it's just 8x minus x squared, so now let's plug in four, so it's going to be eight times four minus four squared and then we'll plug in zero , so it is eight times zero minus zero squared eight times four is thirty-two four squared is sixteen and everything to the right will be zero and thirty-two minus sixteen is sixteen so we get the same answer and it is confirmed like this is how you can calculate the area of ​​the region using this particular definite integral now we can also get the same answer in terms of y instead of in terms of x so we can integrate it from c to d using this formula so remember that x is equals f of y, so what we need to do to get a function of y is we need to solve for x so that we have y equals 8 minus 2x.
I'm going to move this to the left side where it's going to be positive 2x and I'm going to move y to the right side where it's going to be negative y and now we divide everything by two to get x is equal to four minus half of y, so we could say that f of y is equal to four minus half of y. We're going to integrate it from c a d o from 0 to 8. then the area will be the integral from 0 to 8 f of y, which is 4 minus half of y and then d y, so the antiderivative of four will be to be 4y the primitive of and to the first power is and to the second power over two evaluated from zero to eight so this is four and minus one quarter times and squared so let's substitute eight, we will have four times eight minus one quarter times eight squared and then once we plug zero into this expression, everything will be zero, so four times eight is thirty-two and then 8 squared is 64. 64 times 1 4 is 16.
So this 2 will give us the same answer of 16 square units for the area , so regardless of whether we choose to find the area in terms of x or in terms of y, we will get the same answer, let's work on this problem, let's calculate the area of ​​the region bounded by the line y is equal to x and y is equal to x squared and is equals x is basically a straight line that passes through the origin at a 45 degree angle, so y equals x and equals x squared is basically a parabola that opens upward but starts at the origin, so If put those two graphs together, we'll get something that looks like this, so here's the line y equals x and here's y equals x squared on the right side, on the left side, it just looks like this, but we don't have to.
To worry about that, the area of ​​the region bounded by these two curves is right here, so that's the area we need to calculate. What we need to do is determine the intersection points that we can see the first point going to. be at zero, but we have to find the second point to do that, we want to set these two equal to each other, so we want to set x equal to x squared, subtracting both sides by x, we will get that zero equals x squared. minus x and then if we factor the gcf if we factor x we ​​will get this expression using the zero product property we can see that x is equal to zero and x is also equal to one so those are the points of intersection now the top function is our function f of x, which is y equals x, the bottom function is our function g of x, which is x squared, so now we can calculate the area between the two curves using this formula, it is the definite integral from a to b of the function upper f of x minus the lower function g of x, so this will be the integral from 0 to 1. f of x is x g of x is x squared the antiderivative of x will be x squared over 2 and for x squared Using the power rule it's half minus one over three let's multiply half by three over three and one over three by two over two to get common denominators, so this becomes three over six, this becomes two over six and our final answer is one over six , so this is the area between the two curves is one in six square units number three calculate the area of ​​the region bounded by the curves y equals x squared and x equals y squared so let's graph these two functions separately so this is y is equal to x squared is a parabola that opens upward now x is equal to y squared opens to the right if you take the square root of both sides you will get that y is equal to the square root of x plus or minus The top of this function is y equals the positive root x the bottom is y equals the negative square root x Now if we focus on quadrant one where the two curves meet, this is y equals x squared and this is x equals y squared, so now this is the top function which is f of x, which is basically that function x equals y squared, but by solving y we know that the square root of of x is equal to y and remember that f of y is equal to x so be careful, don't confuse them, so we can replace f of x with what is equal to y, which is the square root of x, the bottom function is y is equals x squared and that's our function g of x g of x is also equal to y, it's just a different type of y, so this is x squared and our goal is to calculate the area of ​​the shaded region, so that first we need to find the points of intersection, so let's set f of x equal to g of x, so f of x in terms of x is the square root of x g of x is x squared if we square both sides we will get that x is equal to x to the fourth now this is true when x is zero zero is equal to zero to the fourth and this is true when x is equal to one one is equal to one to the fourth power, so those are the points of intersection now that we have that we can calculate the area using this formula, so the area will be the definite integral from a to b or zero to one of the higher function which is the square root of x, but we will write it as x halved minus the function lower which is three.
The antiderivative of x squared is x cubed over three. Now let's connect one. so this is going to be two over three minus one over three and when we plug in zero we get zero, so the final answer is just one over three, so that's the area between the area of ​​the region bounded by these two curves. so that's the final answer, now let's move on to number four, let's calculate the area of ​​the region bounded by the curves x equals one minus y squared and x equals y squared minus one, so first let's talk about how we can graph this . knowing that x equals y squared looks like this if that's x equals y squared what is x equals y squared minus one?
All we have to do is take this curve and subtract it by one along the x axis so it moves one unit to the left it will start at horizontally one unit to the left, so now that we have the graph of that function, let's focus on the graph of this function now x equals x equals y positive squared opens to the right what about x equals y negative al square? What happens then? So this will open to the left and is reflected over the y axis. Now, what happens if we add a 1 when we subtract one on the graph?
It started at negative one, if we add one, it will start at positive one, so x equals negative and squared plus one, we're going to move one unit to the right, so this is the same as x equals one minus and squared, so what? What we need to do is combine these two graphs so that it looks like this and this will be the area of ​​the region bounded by those two curves, so let's redraw that image, so here we have y squared minus one and here Actually , this was one minus y squared and this one is going to be y squared minus 1. so let me write that down, so this is x equals y squared minus 1 and this one here is x equals one minus y al square so how can we determine the area of ​​the region bounded by these two curves?
First we need to find the points of intersection, so let's set these two expressions equal to each other and, since we have x in terms of y, the points of intersection will be y values, so let's set one minus y squared equals y squared minus one, so let's add y squared to both sides and add one to both sides so they cancel we will get 2 equals 2y squared by dividing both sides by 2 we get 1 equals y squared and then if we take the square root of both sides, we get that 1 is equal to plus or minus 1, which should be written this way, so this is a y value of negative 1 and this is a y value of positive 1. so now that we have the y values ​​of the intersection points we can calculate the area between the two curves using this formula, it will be the integral from c to d, which are the y values ​​of the right function f of y minus the left function g of and then f of y is here f of y is one minus y squared we could see it here instead of the function on the right g of y is the function on the left this is g of y is on the left side for region that is bounded so g of y is y squared minus 1. so now we could use this formula, it will be the integral from c to d c is negative 1 d is positive 1. and then f of y, which is 1 minus y squared minus g of y which is y squared minus 1 d and so first let's simplify this expression, this is one minus y squared minus y squared and then these two negative signs will become a positive one, so one plus one is two negative and squared minus y squared is negative two and squared the antiderivative of 2 is 2y and for 2y squared we have 2y raised to the third over 3, from negative one to one, so let's plug one in, this will be two times one minus two by one raised to the third over three and then if we replace minus 1 we will get this, so this becomes 2minus 2 over 3 and then minus two plus two over three distributing the negative sign will have plus two minus two over three now let's combine the like terms two plus two is negative four two over three minus two over three that's negative four over three so what What we need to do now is get common denominators so four over one I'm going to multiply that by three over three so this becomes twelve over three minus four over three which gives me a final answer of eight over three so this is the area between the two curves.
Now let's work on some more practice problems. Let's calculate the area of ​​the region bounded by the line y is equal to x squared minus 4x and the x axis. squared we know that it opens like this x squared minus 4x is going to be similar, but it's going to shift, so what we have to do is graph this. We need to find the intersections with the x-axis and the vertex. To find the x-intercepts, let's replace y with zero, solve for x, so I'm going to factor x using a zero product property we get an x-intercept of 0 and 4. so let's make a table so that when x is 0 y let 0. when x is 4 and is also 0. now the vertex of a parabola is the midpoint of the x-intercepts, so it will be at 2.
Another way to get the vertex is by using this equation x equals negative b over 2a, let's say if you have a quadratic equation ax squared plus bx plus c then a is the number in front of x squared, which is one b is the number in front of x, which is negative 4. then we have negative b over 2a , so this is positive 4 over 2, which gives us an x ​​value of 2 Now we need to determine the value of y to do that, we can simply plug it into this expression so that y is equal to x times x minus four, if we plug in two, this is two times two minus four, so we get two times negative two, which is negative four so we have enough information, wow, let's go back so we have enough information to graph this function now, so we have an intersection x of zero and four and that negative four we have the vertex at an x ​​value of 2. then the graph is a parabola that opens upwards in this way, our objective is to calculate well the area of ​​the region delimited by that line, technically it should be a curve, not a line, so let's replace that with the word curve and the x-axis, so how can we?
To find the area of ​​that shaded region, we need to identify the top feature and the bottom feature. The top function is the x axis and that is basically the line y is equal to zero, so we can say that the top function is f of x is equal to zero. the lower function g of x is the curve x squared minus 4x and we are going to integrate this from 0 to four, so let's use this formula, it will be the integral from a to b of the upper function f of x minus the lower function g of x so this is going to be the definite integral from zero to four the top function is zero the bottom function is negative well it's negative x squared minus 4x so let's go ahead and distribute the negative sign so that negative multiplied by negative 4x that's going to be positive 4x and then negative x squared, the antiderivative of 4x will be 4x squared over 2. and for x squared it will be x cubed over 3. so this becomes two x squared, so we have two x squared minus x cubed plus three evaluated from zero to four so let's plug in four when we plug in zero all of this will be zero four squared is sixteen times two that's thirty-two four to the third power is sixty-four so we have sixty-four three now we need to get common denominators, let's multiply 32 over one times three over three 32 times 3 is 96 so we have 96 over 3 minus 64 over 3 and 96 minus 64 is 32. so the area of ​​the shaded region is 32 over 3. that's the answer for this problem number six, calculate the area of ​​the region bounded by the equations y equals x squared minus 4x and y equals 6 minus 3x, so this equation that we are familiar with we already know that the intersections with the axis x will be 0 and four and we know that the vertex will be at two minus four, so this graph will open up like this, so it's a rough sketch.
Here we have a linear equation y is equal to 6 minus 3x, so the y-intercept is 6, which should be somewhere here y To determine the x-intercept, we can replace y with 0 and solve for x, so adding 3x to both sides we have 3x equals 6 and 6 divided by 3 is 2. So this line will touch the x axis at 2. so that's what we have and eventually this curve will meet this line, so our goal is to obtain the area of ​​the shaded region. What we need to determine are the intersection points here and here, so we need to calculate the area by taking the top function and subtracting it by the bottom function, so we want the x values ​​of the intersection points, so let's set these two functions equal each other x squared minus 4x, let's set that equal to 6 minus 3x, so let's get started. so let's add 3x to both sides and subtract both sides by 6. then on the left side we will have x squared minus 4x plus 3x, that will be negative x and then we will have negative 6. then we could factor x squared minus x minus six two numbers that are multiplied by negative six but added to the coefficient of the middle minus one is negative three and plus two so we can write this as x minus three times x plus two solving for x we ​​get that x is equal to plus 3 and if you set x plus 2 equal to 0 you will get that x is equal to negative 2. so here is the first point of intersection, this one is at an x ​​value of three and this is at an x ​​value of negative two and the top function that is f of linear 6 minus 3x, the bottom function g of x is the other one which is x squared minus 4x, so now let's calculate the area using this formula, so it will be the integral from a to b, which is negative two to three f of x which is six minus three x minus g of x which is x squared minus four x so we have 6 this is negative 3x minus negative 4x which is negative 3x plus 4x so that's just plus x and then have negative things the antiderivative of 6 will be 6x 4x will be x squared over two and for x squared it will be raised to the third power over three and now let's plug in negative two six times three is eighteen three squared is nine three to the power of three is 27 divided by three, that's nine and then we have negative six times negative two is negative twelve minus two squared is four divided by two is positive two minus two raised to the third power is negative eight with the negative in front which becomes positive eight but divided by three so here we can combine eight, I mean eighteen and minus nine eighteen minus nine is nine and here we have negative twelve plus two, which is negative ten and then plus 8 over 3.
Distributing the negative sign we have 9 plus 9 over 2 and then we have plus 10 minus 8 over 3. so let's first combine nine and ten to get just nineteen and then let's focus on combining those two fractions, let's try to get a common denominator of six, so nineteen over one I'm going to multiply that by six over six nine over two I'm going to multiply that by three over three and eight over three I'm going to multiply that by two over two nineteen times six is ​​one hundred fourteen nine times three is twenty-seven eight times two is sixteen twenty-seven minus sixteen is eleven and one fourteen plus eleven is one twenty-five so this is the final answer this is the area of ​​the shaded region it is 125 over 6 number seven then we want to find the area of ​​the region bounded by these two equations we get a linear equation and a parabolic equation.
Let's start by graphing the parabolic equation so we have is equal to 3y minus 2. So when y is 0, x will be negative 2. That means the x-intercept is negative 2. Now we have a positive coefficient in front of y that is positive 3y and not negative 3y, so we know that we are going to have a linear equation that will increase as x increases as y increases x will increase so it will increase so we want to find the area of ​​the shaded region what we need to do is find the points of intersection first , so let's set 3y minus 2 equals 2y squared minus 4.
So I'm going to take these numbers and move them to the other side, so let's have 0 equals 2y squared, this will be negative 3y on the right side and then it will be positive 2 on the right side so negative 4 plus positive 2 will be negative 2. So now how can we factor this trinomial where the leading coefficient is two to factor it we need to multiply the leading coefficient by the constant term two times negative two is negative four and then we need to find two numbers that multiply to negative four but add to the coefficient of the middle minus three this is going to be negative four and plus one, so we are going to replace the middle term minus three and with negative four and plus one and and then we could factor by grouping, so we're going to remove the gcf in the first two terms and that's 2y and we'll be left with y minus negative 2 and we're going to remove the gcf in the last two terms, which will be simply 1 and then multiplied by y minus 2.
Now these two are equal, so let's factor y minus 2 and we're left with 2y plus 1. This is how we can factor this trinomial if we set y minus 2 to 0, we'll get y is equal to positive 2 and if we set 2 and plus 1 equals 0 and it's going to be negative 1 over 2. so now we have our intersection points and these are values ​​of y minus 1 half and positive 2. so this is equal to c and this is equal to d since everything is in terms of and here, so now we could use this formula to get the area; will be the integral from c to d of the function on the right, which is f of y, minus the function on the left, which is g of y, so relative to the shaded region the function on the right which is here that function is f of and that is the linear equation three and minus two the function on the left that is the curved function that is g of and that has to be the other two and squared minus four, so the area will be c a d or minus half to two f of y, which is three and minus two, and then minus g of y, which is two and squared minus four d y, so we have three and minus. two minus two and squared and then plus 4. so this becomes negative 2y squared plus 3y and then combining these two will be plus 2. the antiderivative of negative two and squared will be negative two and to the third over three for three and is 3y squared over 2 and for 2 it is 2y evaluated from negative one half to 2. so now let's connect two, this will be negative two times two raised to the third power over three plus three times two squared over two plus two times two and then let's plug in negative one half so that it's negative two times negative one half to the third power over three plus three times negative one half squared over two plus two times negative half two to the third power is eight times negative two that's negative sixteen divided by three two squared is four divided by two is two times three which gives us six and then two times two is four minus one half raised to the third power is negative one over eight minus one over eight times negative two is positive one over four divided by three that becomes positive one over twelve minus one half squared is positive one quarter divided by two that's one over eight times three that's positive three over eight and then two times minus half is minus one so we have negative 16 over three six plus four is ten distributing the negative sign we have negative one over twelve minus three over eight and then this is plus one so what we have to do is get a common denominator of three twelve and eight the best number is 24. 24 is a multiple of 3 12 and 8. but first let's combine 10 and 1, which is 11. we will multiply this by 24 over 24 and then minus 16 over 3, we will multiply it by 8 over 8. negative 1 over 12 we will multiply by 2 over and minus three over eight we will multiply by three over three to get a common denominator of 24.
So we have 11 times 24, which is 264 16 times 8 is 128 2 times 1 is 2 three times three is nine so two sixty and four minus one twenty-eight is one hundred thirty-six and minus two minus nine is minus eleven 136 minus 11 is 125. so the answer is 125 over 24 square units, so that is the area of ​​the region bounded by the two curves or the two equations that