Area Between Curves: Videos & Practice Problems

To find the area between two curves, set up a definite integral from the lower to the upper bounds, which are determined by the intersection points of the functions. The area can be calculated by integrating the top function minus the bottom function. If the functions intersect within the interval, split the integral into two parts, adjusting the top and bottom functions accordingly. This method ensures accurate area calculations, even when functions switch positions across the interval.

Finding Area Between Curves on a Given Interval

Earlier we learned how to find the area underneath the curve of a function by setting up a definite integral. But our area calculations can get a bit more complicated than this, because we won't just be asked to find the area underneath a single curve, but also the area in between two curves. Now luckily, because of our knowledge of definite integrals and the area that they represent, I think you'll find this process to be pretty straightforward. And I'm going to walk you through exactly how to set it up here. So let's go ahead and get started by jumping right into our example.

Here, we're asked to find the area of the region in between these two functions, f of x and g of x, that we see on our graph here, specifically from x equals zero to x equals one. Now whenever we're faced with these sorts of problems, finding the area between two curves, we always want to start by first graphing our functions and then shading the area in between them so that we can fully visualize what's going on with our problem and avoid making any errors. Now we already have our two functions on our graph here and we know that the area that we want to find in between them is right here specifically on that interval from zero to one. So how exactly can we go about finding this area? Well, thinking back to finding the area underneath a single curve by setting up a definite integral, let's go ahead and break this down a little bit.

If I just wanted to find the area underneath this function on the top, this f of x, I could go ahead and set up a definite integral in order to do so. And I could do the same exact thing for my function g of x on the bottom here, setting up a definite integral from zero to one to find the area underneath that curve there. But, in order to find the area in between them, what do I need to do? Well, looking at these two areas here, if I find the area underneath that function f and then I subtract off the area underneath my function on the bottom here, g, that will leave me with the exact area that we're looking for in between the two curves of our function. So that's exactly what we want to do here.

We want to go ahead and integrate that top function minus that bottom function. Now if we look at this in integral form here, we wanna take the definite integral on our specific interval from a to b of f(x)dx minus the definite integral on that same interval of g(x)dx, so subtracting off that integral of the bottom function from the top function. Now based on our integral rules, we can combine this all into one integral, the integral from a to b of f(x)−g(x)dx. And this would give us the area in between our two functions, f of x and g of x. So let's actually go ahead and set this integral up for two functions here so that we can actually find the area that we're looking for.

So this specific area is gonna be from zero to one, setting up the bounds on my definite integral here. We were given those bounds originally in our problem, and we can see based on the area that we shaded that our bounds should be zero and one. Now we're going to take that top function, so here that's four minus x squared, and then we're going to subtract off that bottom function. Here that's two x plus one. Then integrating here with respect to x.

So we have our integral fully set up and now we just need to evaluate it to actually find that area. Now I'm going to go ahead and distribute this negative into my parenthesis here that will give me negative two and negative one. And I'm going to rewrite my integrand in standard form and go ahead and combine like terms. So this is the definite integral from zero to one of negative x squared, writing this in descending order of power, minus 2x. And then if I have a positive four and a negative one, having distributed that negative, that will leave me with a positive three.

Then, of course, integrating here with respect to x. So now that we've simplified this, we can go ahead and apply the power rule to get our antiderivatives here of each of these terms. So doing that using the power rule on that term negative x squared, that's going to give me negative one third x cubed and then minus two times one half x squared applying our power rule there as well and then finally plus three x and bounded from zero to one. So now we can go ahead and apply those bounds to get an actual numerical answer here using the fundamental theorem of calculus plugging in that upper bound so negative one third one cubed. I'm going to go ahead and cancel these twos here.

So I just have a minus one squared still plugging in that upper bound, plus three times one, then subtracting off having plugged in that lower bound. If I plug zero into each of these terms, they will all work out to be zero. So we are just subtracting off zero here. And if we work this out algebraically, we will end up getting a final answer here of five thirds. This five thirds gives me the area in between the two curves of my function.

Having set up my definite integral of that top function minus my bottom function. Now it's important to note here that this method will work even if our functions are below the x axis. So we're still gonna get the correct area regardless of where those functions are located. Now, we're going to continue getting practice finding the area in between curves coming up in the next couple of videos. Let us know if you have any questions and I'll see you there.

Calculate the area of the shaded region between the 2 functions from $x=0$ to $x=9$

• A. 7
• B. 10
• C. 4.5
• D. 445

Answer: (C)

Calculate the area of the shaded region between $f\left(x\right)$ & $g\left(x\right)$ contained between $x=-4$ & $x=-2$.

• A. 8.17
• B. 4.17
• C. -12.17
• D. 0.50

Answer: (A)

Sketch the region bounded by $f\left(x\right)=-{(x-2)}^2+5$ & $g\left(x\right)=4x$ on the interval $[0,1]$. Then set up an integral to represent the region's area and evaluate.

• A. Integral = ${\int }_0^1(-x^2+9)dx$; Area: $0.33$
• B. Integral = ${\int }_0^1-x^2+1dx$; Area = $0.67$
• C. Integral = ;${\int }_0^4(-x^2+9)dx$ Area: $21.33$
• D. Integral = $\int_0^1-x^2+1dx$; Area = $22.67$

Answer: (B)

Shade the region bounded by $f\left(x\right)=\sin x$ & $g\left(x\right)=\cos 2x$ on the interval $[\frac{\pi }{6},\frac{5\pi }{6}]$. Then set up an integral to represent the region's area.

• A. $A={\int }_{\frac{\pi }{6}}^{\frac{5\pi }{6}}(\cos 2x-\sin x)dx$
• B. $A={\int }_{\frac{\pi }{6}}^{\frac{5\pi }{6}}(\cos 2x-\sin x)dx$
• C. $A={\int }_{\frac{\pi }{6}}^{\frac{5\pi }{6}}(\sin x-\cos 2x)dx$
• D. $A={\int }_{\frac{\pi }{6}}^{\frac{5\pi }{6}}(\sin x-\cos 2x)dx$

Answer: (C)

Finding Area Between Curves on a Given Interval

In this problem, we're asked to find the area between the two functions f(x) = |x + 2| and g(x) = x² on the interval from zero to two. So let's go ahead and take a look at what region we're trying to find the area for. Looking at our graph here, we can see that we have this function f(x) on top and our function g(x) on the bottom. Now, we're asked to find the area from x = 0 to x = 2. So if I go ahead and shade that area in here, that is this yellow shaded area.

How can we go about finding this area? Well, we know that we need to set up a definite integral of our top function minus our bottom function. Based on what's given to us in the problem, we know that we want to bound this integral from zero to two. Then I want to take my function that's on top, which is f(x), that absolute value of x + 2, and then subtract off that bottom function. In this case, that g(x) is equal to x².

So subtracting off that x² and then integrating with respect to x here. Now, something that's important to note here, because we're working with an absolute value function, looking at the bounds that we're working with from zero to positive two, all of these x-values are going to be positive. So, I really don't have to worry about those absolute value lines at all because all of these values are already positive. So, I can actually rewrite this integral as the integral from zero to two of (x + 2 - x²) dx. Now, I can just go ahead and evaluate that integral the way that I know how to find this area.

I'm going to go ahead and rewrite this polynomial in standard form, putting that x² upfront and then a + x + 2, bounded from zero to two. Now, going ahead and finding our antiderivative here, the antiderivative of -x², this is going to be -13x³, and then + 12x² + 2x bounded from zero to two. Now we want to go ahead and apply our bounds using the fundamental theorem of calculus. Plugging in two here, this is going to be -13 times 2³ + 12 times 2² + 2 times 2. Lots of twos here.

Now I want to subtract off that lower bound plugged in, but because all of these values have x in them or all of these terms have x in them, all of them are just going to be zero. So we're really just subtracting off zero there, and there's no need to write that all out. So now we just have to do some algebra, multiplying this out, getting all of our values. So 2³ is eight. So this first term is -8/3 and then 2² is four divided by two is two, so +2 and then +4.

Now I want to go ahead and get all of these with a common denominator here. So I'm going to make that three, so -8. And then if I make this have a denominator of three, this is going to be +6/3. And then if I make this four have a denominator of three, that's +12/3. Now if I have +6, +12, and -8 here, that's going to be +18 and -8.

That gives us 10/3. So as a fraction, this area is 10/3, this shaded area right here. Now if you want to report this as a decimal, 10/3 is going to be about 3.33. But of course, that fraction 10/3 gives us a more exact answer. So here we can see that even though we had an absolute value, we didn't have to worry about those absolute value lines because we took a look at the bounds of our integral and where we're actually working with that function, only for positive values.

If you have any questions here, feel free to let us know, and let's keep going.

Finding Area When Bounds Are Not Given

We just learned how to find the area between two function curves by setting up a definite integral bounded from a to b. Now, in all the examples we've worked through so far, these bounds were explicitly given to us, but this will not always be the case. So, how exactly can we go about setting up this integral and finding the area between two curves when those bounds are not given? Well, luckily, we can do this rather easily by just using basic algebra. Now, we're going to walk through this process together by jumping right into our example here where we're asked to find the area between our two functions y=4-x2 and y=2x+1.

Now looking at these two functions on our graph here, I can see that the area between these two curves is contained between the points where these two functions actually intersect each other. And that's exactly what our bounds are going to be. Whenever the bounds are not given to us explicitly in our problem, we want to go ahead and use our intersection points to bound our integral and ultimately find the area between our two functions. Now, this is where algebra skills come in because, in order to find these intersection points, we need to set our functions equal to each other and then solve for x. So that's exactly what we want to do here.

We want to take f(x) and set it equal to g(x). So for this problem specifically, we're taking that 4-x2 and setting it equal to 2x+1. From here, we want to go ahead and solve for x. So the first thing that I want to do is get everything all on one side. So, I'm going to go ahead and add x2 to both sides to get it to cancel on this left side and also subtract four.

So that leaves me with a zero on that left-hand side and then 0=2x+1+x2-4. Now I'm going to go ahead and write this in standard form, doing some rearranging here and also combining these like terms, my constants. So this gives me x2+2x-3=0. So here I can see that we have a quadratic equation that we need to solve. Now, there are multiple ways that we could go about solving this quadratic equation, but the most simple way to approach this is going to be to factor.

We're looking for two factors that multiply to negative three and add to positive two. So those two factors are positive three and negative one. So this quadratic equation factors to (x+3)(x-1), and we know that this is still equal to zero. Now because this is equal to zero, we can take each of these individual factors and set them equal to zero to ultimately find these values of x. So taking that x+3, setting that equal to zero, if I subtract three on both sides of this equation here, I'll cancel that three out on the left just leaving me with x and then x will be equal to negative three.

So that's my first intersection point. Now doing the same thing with my other factor here, x-1 is equal to zero, If I add one to both sides here, that gives me x equals positive one and that is my other intersection point. So these are the intersection points that we can see on our graph here that our area is contained between. Now let's not lose the plot here because remember, in finding these intersection points, we were really finding the bounds to our integral. So now that we have these bounds, we can actually set up this integral and find the area between our two functions.

So, setting up our area integral here, having found those bounds, this is going to be the integral from negative three to positive one. Then I want to go ahead and take my top function minus my bottom function. Looking at my graph over here, I can see that my top function is that 4-x2 and that bottom function is that 2x+1. So top function 4-x2, then subtracting our bottom function 2x+1, integrating with respect to x. So with my area integral set up, we just need to do some simplification and ultimately evaluate our integral.

So going ahead and cleaning up that integrand here, I'm going to go ahead and combine like terms and distribute that negative to make this the integral from negative three to positive one of -x2-2x+3 integrated with respect to x. Now we can go ahead and evaluate this, finding our antiderivative. That first term using the power rule, I'm going to get -13x3 and then -x2+3x, bounded from negative three to one. Now applying our fundamental theorem of calculus here, we know that we need to plug in that top bound to all of these terms and then subtract off that bottom bound plugged in. You can go ahead and pause here and try that on your own.

Once you get through that algebra, you should end up getting a final answer here of 323. So this is the area between our two functions that we see here, having found our bounds by using our intersection points. Now we're going to continue getting practice with this type of problem coming up next. I'll see you there.

Find the area between $f\left(x\right)=x^2-4$ & $g\left(x\right)=-x^2+4$.

• A. 0
• B. 10.67
• C. 32
• D. 21.33

Answer: (D)

Find the area of the shaded region between $f\left(x\right)=x^4-x^2$ & $g\left(x\right)=3x^2$.

• A. 8.53
• B. 17.07
• C. 34.13
• D. 4.27

Answer: (A)

Find the area of the shaded region ONLY that lies between the line $y=1$ & $f\left(x\right)=\sin 2x$.

• A. 2.43
• B. 1.73
• C. $\pi$
• D. $2\pi$

Answer: (C)

Finding Area Between Curves that Cross on the Interval

Up to this point, we've been finding the area between two curves on an interval where it's super clear which function is on the top and which function is on the bottom. But what if we were asked to find the area between two curves on an interval where that's not the case? Where a function may start out on top but then become the bottom function as it intersects the other one. How exactly could we go about finding all of that area? Well, luckily, we can actually do this by splitting this up into two separate integrals that we can set up and evaluate the same exact way that we've been doing.

Now we're going to walk through this process together here. So as usual, let's jump right into our example. We are asked to set up an integral to represent the area between f(x) = 4 - x2 and g(x) = 2x + 1 on the interval from zero to two. Now taking a look at my graph here, on my interval from zero to one, we can clearly see that this function f(x) is on top. So we know exactly how to set up an integral in order to find this area.

But then on my interval from one to two f(x) is no longer the function that's on top. We can see that our function g(x) is on top now, meaning that we would need to set up this integral differently in order to find this blue shaded area. And that's exactly what we're going to do. Whenever we want to find the area between two curves that intersect within a given interval, we want to go ahead and set up two integrals and just add them together.

So here it's important to note that both of these integrals are still going to be our top function minus our bottom function, but our top and bottom functions are going to switch after that intersection point, which is exactly what we see happen on our graph here. So let's go ahead and set up these integrals in order to find this area. Taking a look at this first area on my interval from zero to one, I see that function f(x) on top there. So I want to take the integral from zero to one of that top function f(x) = 4 - x2, minus my bottom function g(x) = 2x + 1 integrated with respect to x. Now looking at the second half of my interval from one to two, f(x) is no longer my top function.

So instead of taking f(x) minus g(x), we're going to take g(x) and subtract f(x) because now g(x) is that top function, f(x) is that bottom function. So adding that integral together here from one to two of that top function, in this case now 2x + 1, minus that bottom function 4 - x2 integrated with respect to x. Now these integrals together represent that total area between my two curves on this interval from zero to two. So from here, we could go ahead and evaluate these integrals and get a numerical answer for our area. You should go ahead and try to finish up that process on your own, and you can check your answer with me in the next video.

I'll see you there as we continue to practice.

Finding Area Between Curves that Cross on the Interval Example 2

If you haven't already watched the previous video where we show exactly how to set up this interval to find the area between these two functions that cross on our interval, go ahead and do that before coming back here and checking your answer with me. For those of you that have already done that and set up this interval, we're going to continue working through it together. So here with this problem, we were asked to calculate the area between our two functions f(x) and g(x) on the interval from zero to two. Now based on our graph here, we can see that in the middle of this interval, these two functions switch which one is on the top and which one is on the bottom, meaning we had to set up two separate integrals that we are adding together in order to get each of these individual areas. This first integral from zero to one of \(4 - x^2 - 2x + 1 \, dx\), this represents this first chunk of area here from zero to one.

Then as these curves cross each other and g(x) becomes my function on top, we add in this additional integral from one to two of \(2x + 1 - 4 - x^2 \, dx\) in order to get this second chunk of area here. We're going to go ahead and finish working through this problem and finding what this area is. So starting with this first integral, I'm going to go ahead and do some simplifying based on what's in my integrand here. So writing this integral in standard form: the integral from zero to one of \(-x^2 - 2x + 3 \, dx\). Then for that second integral from one to two, again writing this in standard form, this is going to be \(x^2 + 2x - 3 \, dx\). We can see that what's in these two integrals is almost identical. They just have flipped signs, which makes sense because we flipped which function we were subtracting and which one was it was being subtracted from. Let's go ahead and find these antiderivatives here.

For this first integral, this is going to be \(\frac{-1}{3}x^3 - x^2 + 3x\) bounded from zero to one. And then adding in this next integral here, same exact antiderivative, just with switched signs. So this is \(\frac{1}{3}x^3 + x^2 - 3x\) bounded from one to two. At this point, we just have to apply our fundamental theorem of calculus and plug these bounds in. Starting with this first integral or this first antiderivative plugging in that upper bound, this is going to be \(\frac{-1}{3}(1)^3 - (1)^2 + 3(1)\). So that's that upper bound. If I subtract off that lower bound plugged in since all of these terms have an x in them multiplying there, all of these terms are going to end up being zero. So plugging in that lower bound, we're just really subtracting off zero and this represents that first integral. Then we're adding this together with these two bounds plugged into this antiderivative here.

So this is \( \frac{1}{3}(2)^3 + (2)^2 - 3(2)\) for the upper bound. And then we are subtracting off this lower bound plugged in \( \frac{1}{3}(1)^3 + (1)^2 - 3(1)\). Now we just have algebra to do from here. If you wanna go ahead at this point and plug this all into your calculator at once to get your final answer, you can. Just be very careful with your parentheses placement, but I'm going to go ahead and continue working through this algebra by hand as I would have to do if I wasn't allowed a calculator on a homework or an exam. So let's continue on here.

Looking at this first set of terms here, we don't have to worry about that zero but we have \( \frac{-1}{3}(1)^3\) that is just \( \frac{-1}{3}\) and then \(-1\) and then \(+3\). That's that first set of terms there. Then looking at this next one it gets a little bit messier here but if I take two cubes that is eight. So this first term here is \( \frac{8}{3}\) and then the second term, \(2^2\) is four. \( \frac{8}{3} + 4 - 3 \times 2\), that's going to be minus six. So this leaves us with \( \frac{12}{3} \) or just four. So our final answer for this area here between these two functions f(x) and g(x) from zero to two is four.

Feel free to let us know if you have any questions here, and let's continue practicing in the next video.

Find the shaded area between $f\left(x\right)=x^3+2x^2$ & $g\left(x\right)=x+2$.

• A. 2.25
• B. 3.08
• C. 0.42
• D. 2.67

Answer: (B)

Find the area of the shaded region between $f\left(x\right)=\sin 2x$ & $g\left(x\right)=2\sin x$ from $x=0$ to $x=2\pi$.

• A. 8
• B. 0
• C. 4
• D. 6

Answer: (A)

Finding Area Between Curves that Cross on the Interval Example 3

If you haven't already, go ahead and try this problem out on your own before checking back in with me. Alright. Here, we're asked to set up a definite integral to represent each of the three shaded areas that we see on our graph here. So let's start with that first area shaded in yellow down here, area A. Now we can see that this is a rather small sliver of area here, and it's contained with the function g of x on top and my function f of x on the bottom.

And we can see that this boundary further goes from negative six until these two functions actually intersect each other at negative five. So this is a rather small interval. Thus, this area is going to be the integral from negative six to negative five of our function that's on top. Again, g of x here, we can see that labeled on this side, g⁡x minus our function that is on the bottom here. We can see this function in red labeled f⁡x.

So g⁡x-f⁡x. Remember, that's our top function minus our bottom function, of course, integrated with respect to x here. So that's our integral that represents our yellow shaded area. So let's move on to our next piece. Here we have area B shaded in this sort of light red pinkish color, and we can see that this is the area contained underneath my function f of x in between that function and the x-axis.

So for this particular integral, we don't have to worry about subtracting anything off here. Now since we've been working with the area in between curves here, you may be thinking okay, well if I'm finding the area between this function and the x-axis, I need to subtract off the x-axis. But that's just zero, right? So you would be subtracting f⁡x-0, which is just f⁡x. Remember that the area underneath a curve is the area between that curve and the x-axis, So we just need a definite integral here, which would be from negative four to that origin point zero of that function f⁡xd⁡x. And that's all.

We're not looking at the area in between curves here, just the area underneath a single curve. Now let's move on to our final area here. This is area C shaded here in green, and this area is in two separate pieces, so we need to set up two different integrals because we can see that our function g of x starts out on top but then becomes the bottom function as these two functions intersect each other. So let's start with that first section of area C. So in order to find this area, we want the definite integral here from the origin until these two functions intersect each other at x equals seven.

So from zero to seven of the function on the top, which here is my function in blue g⁡x minus my function on the bottom here in red f⁡x, then integrated with respect to x. Now that's only this larger area. We still need to add on this smaller sliver of area here. So we're adding another definite integral here from seven, that intersection point, until this interval ends here at nine, so seven to nine, of what is now our top function f⁡x minus what is now our bottom function g⁡x.

We can see that those top and bottom functions flipped between these two integrals, and then integrated with respect to x. So here we have our three integrals representing our three shaded areas. Now if you have any questions here, feel free to let us know in the comments, and I'll see you in the next one.

Here’s what students ask on this topic:

How do you find the area between two curves using definite integrals?

To find the area between two curves, you first need to identify the functions and the interval over which you want to calculate the area. The interval is typically determined by the intersection points of the two functions. Once you have the interval, set up a definite integral from the lower bound to the upper bound. The integrand should be the top function minus the bottom function. This means you subtract the function that lies below from the function that lies above within the given interval. If the functions intersect within the interval, you may need to split the integral into two parts, adjusting the top and bottom functions accordingly. This method ensures you accurately calculate the area between the curves.

What do you do if the bounds for finding the area between curves are not given?

If the bounds are not explicitly given, you can determine them by finding the intersection points of the two functions. To do this, set the functions equal to each other and solve for the variable, typically x. These solutions will give you the points of intersection, which serve as the bounds for your definite integral. Once you have these bounds, you can set up the integral of the top function minus the bottom function over the interval defined by these intersection points. This approach allows you to find the area between the curves even when the bounds are not initially provided.

How do you handle finding the area between curves when the top and bottom functions switch?

When the top and bottom functions switch within the interval, you need to split the integral into two separate parts. First, identify the point where the functions intersect and switch positions. Set up one integral from the start of the interval to the intersection point, using the initial top and bottom functions. Then, set up a second integral from the intersection point to the end of the interval, using the new top and bottom functions. Evaluate both integrals separately and add their results to find the total area between the curves. This method ensures you account for the change in positions of the functions.

Can the method for finding the area between curves be used if the functions are below the x-axis?

Yes, the method for finding the area between curves can be used even if the functions are below the x-axis. The key is to focus on the relative positions of the functions to each other, rather than their positions relative to the x-axis. By integrating the top function minus the bottom function over the specified interval, you will obtain the correct area between the curves, regardless of whether they are above or below the x-axis. This approach ensures that the calculated area is always positive and accurately represents the space between the two curves.

What is the role of algebra in finding the area between curves?

Algebra plays a crucial role in finding the area between curves, especially when determining the bounds of integration. When the bounds are not given, algebra is used to find the intersection points of the functions by setting them equal to each other and solving for the variable. This process often involves solving quadratic equations, which may require factoring, using the quadratic formula, or other algebraic techniques. Additionally, algebra is used to simplify the integrand by combining like terms and distributing negatives, ensuring the integral is set up correctly for evaluation. Mastery of algebraic skills is essential for accurately finding the area between curves.