Sketch the region bounded by f ( x ) = − ( x − 2 ) 2 + 5 \textcolor{blue}{f\left(x\right)=-\left(x-2\right)^2+5} f ( x ) = − ( x − 2 ) 2 + 5 & g ( x ) = 4 x \textcolor{orange}{g\left(x\right)=4x} g ( x ) = 4 x on the interval [ 0 , 1 ] \left\lbrack0,1\right\rbrack [ 0 , 1 ] . Then set up an integral to represent the region's area and evaluate.

Integral = ∫ 0 1 − x 2 + 1 d x \int_0^1-x^2+1dx ∫ 0 1 − x 2 + 1 d x ; Area = 0.67 0.67 0.67

Integral = ∫ 0 1 ( − x 2 + 9 ) d x \int_0^1\left(-x^2+9\right)dx ∫ 0 1 ( − x 2 + 9 ) d x ; Area: 0.33 0.33 0.33

Integral = ; ∫ 0 4 ( − x 2 + 9 ) d x \int_0^4\left(-x^2+9\right)dx ∫ 0 4 ( − x 2 + 9 ) d x Area: 21.33 21.33 21.33

Integral = ∫ 0 1 − x 2 + 1 d x \int_0^1-x^2+1dx ∫ 0 1 − x 2 + 1 d x ; Area = 22.67 22.67 22.67

Shade the region bounded by f ( x ) = l o g 2 x + 4 \textcolor{blue}{f\left(x\right)=log_2x+4} & g ( x ) = e x − 4 \textcolor{orange}{g\left(x\right)=e^{x-4}} on the interval [1,4]. Then set up an integral to represent the region's area.

A = ∫ 1 4 ( l o g 2 x + 4 − e x − 4 ) d x A=\int_1^4\left(log_2x+4-e^{x-4}\right)dx A = ∫ 1 4 ( l o g 2 x + 4 − e x − 4 ) d x

A = ∫ 1 4 ( l o g 2 x + 4 + e x − 4 ) d x A=\int_1^4\left(log_2x+4+e^{x-4}\right)dx A = ∫ 1 4 ( l o g 2 x + 4 + e x − 4 ) d x

9. Graphical Applications of Integrals

Area Between Curves

9. Graphical Applications of Integrals

Area Between Curves: Videos & Practice Problems

Video Lessons Worksheet

Topic summary

To find the area between two curves, first identify the intersection points by setting the functions equal to each other. Use definite integrals to calculate the area, integrating the top function minus the bottom function. If the top and bottom functions switch within the interval, split the integral into two parts. The area can be expressed as: $\int_{a}^{b} (f (x) - g (x)) d x$ .

concept

Finding Area Between Curves on a Given Interval

Video duration:

Video transcript

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 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 between these two functions, $f(x)$ and $g(x)$, that we see on our graph here, specifically from $x = 0$ to $x = 1$. 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 between them so that we can fully visualize what's going on with our problem and avoid making any errors. We already have our two functions on our graph here and we know that the area that we want to find 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 bit.

If I just wanted to find the area underneath this function on the top, this $f(x)$, 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 $g(x)$ 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 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 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 want to take the definite integral on our specific interval from $a$ to $b$: ∫ a b f ⁡ ⁡ x - g ⁡ ⁡ x dx 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, ∫ a b f ⁡ ⁡ x - g ⁡ ⁡ x dx . This would give us the area in between our two functions, $f(x)$ and $g(x)$. 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.

This specific area is 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 $4 - x^2$, and then we're going to subtract off that bottom function. Here that's $2x + 1$. 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, which 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 $-x^2 - 2x + 3$.

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 $-x^2$, that's going to give me $-\frac{1}{3}x^3$, then $-x^2$ becomes $-\frac{1}{2}x^2$, applying our power rule there as well, and then $+3x$, 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 $-\frac{1}{3}(1)^3$. I'm going to go ahead and cancel these twos here.

So I just have $-1x^2$ still plugging in that upper bound, $+3$, 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 $\frac{5}{3}$. This $\frac{5}{3}$ gives me the area between the two curves of my function.

Having set up my definite integral of that top function minus the 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 going to 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.

Problem

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

4.5

445

Problem

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

8.17

4.17

-12.17

0.50

Problem

Sketch the region bounded by $\textcolor{blue}{f\left(x\right)=-\left(x-2\right)^2+5}$ & $\textcolor{orange}{g\left(x\right)=4x}$ on the interval $\left\lbrack0,1\right\rbrack$ . Then set up an integral to represent the region's area and evaluate.

Graph showing the area between two curves, f(x) in blue and g(x) in orange, highlighted in yellow.

Integral = $\int_0^1\left(-x^2+9\right)dx$ ; Area: $0.33$

Graph showing two curves, f(x) in blue and g(x) in orange, with the area between them shaded in yellow.

Integral = $\int_0^1-x^2+1dx$ ; Area = $0.67$

Graph showing the area between two curves, f(x) in blue and g(x) in orange, highlighted in yellow on the interval [0,1].

Integral = ; $\int_0^4\left(-x^2+9\right)dx$ Area: $21.33$

Graph showing two curves, f(x) in blue and g(x) in orange, with the area between them shaded in yellow.

Integral = $\int_0^1-x^2+1dx$ ; Area = $22.67$

Problem

Shade the region bounded by $f (x) = l o g_{2} x + 4$ & $g (x) = e^{x - 4}$ on the interval [1,4]. Then set up an integral to represent the region's area.

Graph showing the area between the curves f(x) and g(x) shaded in yellow on the interval [1,4].

$A=\int_1^4\left(log_2x+4-e^{x-4}\right)dx$

Graph showing the area between the curves f(x) and g(x) shaded in yellow, bounded on the interval [1,4].

$A=\int_1^4\left(log_2x+4-e^{x-4}\right)dx$

Graph showing the area between the curves f(x) and g(x) shaded in yellow, with labeled axes and functions.

$A=\int_1^4\left(log_2x+4+e^{x-4}\right)dx$

Graph showing the area between the curves f(x) and g(x) shaded in yellow on the interval [1,4].

$A=\int_1^4\left(log_2x+4+e^{x-4}\right)dx$

example

Finding Area Between Curves on a Given Interval

Video duration:

Video transcript

In this problem, we're asked to find the area between the two functions $ f(x) $ equals the absolute value of $ x + 2 $ and $ g(x) $ equals $ x^2 $ on the interval from zero to two. Let's take a look at the region we're trying to find the area for. Observing our graph, $ f(x) $ is on top and $ g(x) $ is on the bottom from $ x = 0 $ to $ x = 2 $. The area we are focusing on is the yellow shaded area shown.

To find this area, we'll set up a definite integral of our top function minus our bottom function. For this problem, the integral is bound from zero to two. We subtract the bottom function, $ g(x) = x^2 $, from the top function, $ f(x) = |x + 2| $.

Since we're dealing with an absolute value function and the $ x $ values from zero to two are all positive, there's no need to consider negative values. We can rewrite the integral as $ \int_0^2 (x + 2 - x^2) \, dx $. Now, we can proceed to evaluate the integral.

Rewrite the polynomial in standard form: $ -(x^2 - x - 2) $, from zero to two. The antiderivative of $ -x^2 $ is $ -\frac{1}{3}x^3 $, of $ x $ is $ \frac{1}{2}x^2 $, and of 2 is $ 2x $. These are evaluated from zero to two, implementing the Fundamental Theorem of Calculus.

After plugging in the bounds, the integral calculation simplifies as follows:

$ -\frac{1}{3}(2^3) + \frac{1}{2}(2^2) + 2(2) $
Calculating each, we get $ -\frac{8}{3} + 2 + 4 $
To simplify, we convert all to a common denominator, resulting in $ -\frac{8}{3} + \frac{6}{3} + \frac{12}{3} $
This results in $ \frac{10}{3} $

Thus, the area of the shaded region is $ \frac{10}{3} $ square units, or approximately 3.33 when converted to a decimal. This fraction provides a more exact representation of the area.

If you have any questions about this process, feel free to ask. Let's continue with our learning!

concept

Finding Area When Bounds Are Not Given

Video duration:

Video transcript

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 of 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 equals four minus x² and y equals 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 four minus 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 zero is equal to 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 is equal to zero. 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 four minus x² and that bottom function is that 2x + 1. So top function four minus 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 -1/3 x³ 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 32/3. 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.

Problem

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

10.67

21.33

Problem

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

8.53

17.07

34.13

4.27

Problem

Find the area of the shaded region shown in the first quadrant between $f (x) = \frac{1}{x}$ & $g (x) = - \frac{1}{4} x + \frac{5}{4} .$

4.239

1.386

0.489

concept

Finding Area Between Curves that Cross on the Interval

Video duration:

Video transcript

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. Where we're asked to set up an integral to represent the area between $ f(x) = 4 - x^2 $ 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 - x^2 $, 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) - 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 - x^2 $ 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.

example

Finding Area Between Curves that Cross on the Interval Example 2

Video duration:

Video transcript

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 of x and g of 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 $ \int_0^1 (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 of x becomes my function on top, we add in this additional integral from one to two of $ \int_1^2 (2x + 1 - 4 + x^2) \, dx $ in order to get this second chunk of area here. So we're going to go ahead and finish working through this problem and finding what this area is. 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 in standard form, this is 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 $. So 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 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, with the same exact antiderivative, just with switched signs. So this is $ \frac{1}{3}x^3 + x^2 - 3x $ bounded from one to two. So at this point, we just have to apply our fundamental theorem of calculus and plug these bounds in. This can become a bit complex, so it's important to stay organized when doing this work.

So, starting with this first antiderivative, plugging in that upper bound, this is going to be $ -\frac{1}{3} \times 1^3 - 1^2 + 3 \times 1 $. 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}\times 2^3 + 2^2 - 3 \times 2 $. That's my upper bound. And then we are subtracting off here that lower bound plugged in. That's going to be $ \frac{1}{3} \times 1^3 + 1^2 - 3 \times 1 $. Now we just have algebra to do from here.

If you want to 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.

We need to combine like terms and simplify, considering the negative and positive values to handle addition correctly. After handling fractions and combining terms, we can obtain the final solution, which is $4$. This is the area between these two functions f of x and g of x from the interval zero to two. Feel free to let us know if you have any questions here, and let's continue practicing in the next video.

Problem

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

2.25

3.08

0.42

2.67

Problem

Find the area of the shaded region between $f (x) = \frac{1}{x}$ & $g (x) = x$ from $x equals 0.5$ to $x equals 4$ .

-5.796

6.432

0.329

0.557

example

Finding Area Between Curves that Cross on the Interval Example 3

Video duration:

Video transcript

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. We can see that this is a rather small sliver of area here and it's contained within the function g(x) on top and my function f(x) on the bottom.

And we can see that this boundary goes from negative six until these two functions actually intersect each other at negative five. So this is a rather small interval. So this area is going to be the integral from negative six to negative five of our function that's on top. Again, g(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) minus 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. 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(x) in between that function and the x-axis.

So for this particular integral we don't have to worry about subtracting anything 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) minus zero, 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(x) dx. And that's all.

Now, let's move on to our final area here. This is area C shaded here in green and this is two separate pieces of area so we need to set up two different integrals because we can see that our function g(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. In order to find this area, we want the definite integral here from the origin until these two functions intersect each other at x=7.

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). So f(x) minus g(x).

We can see that those top and bottom functions flipped between these two integrals and then integrated with respect to x. 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, first identify the intersection points of the curves to establish the bounds for integration. Once the bounds are determined, set up the integral as the top function minus the bottom function over the interval. The integral is expressed as:

\int a b (f (x) - g (x)) d x

If the functions switch positions within the interval, split the integral into two parts, each with its respective top and bottom functions. Evaluate the integrals to find the area accurately.

Created using AI

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

If the bounds for integration are not given, you can find them by determining the intersection points of the two curves. Set the functions equal to each other and solve for $x$ to find these points. These intersection points will serve as the bounds for your integral. Once you have the bounds, set up the integral as the top function minus the bottom function over the interval. This method ensures you accurately calculate the area between the curves.

Created using AI

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

When the top and bottom functions switch within the interval, split the integral into two separate integrals at the intersection point where the switch occurs. For each segment, set up the integral as the top function minus the bottom function. Evaluate each integral separately and then sum the results to find the total area. This approach ensures that you account for the change in position of the functions and accurately calculate the area between them.

Created using AI

Can you find the area between curves if the functions are below the x-axis?

Yes, you can find the area between curves even if the functions are below the x-axis. The process remains the same: identify the intersection points to establish the bounds, and set up the integral as the top function minus the bottom function. The position of the functions relative to the x-axis does not affect the calculation of the area between them, as the integral accounts for the vertical distance between the curves.

Created using AI

What is the significance of the intersection points when finding the area between curves?

The intersection points are crucial when finding the area between curves because they determine the bounds for the integral. These points indicate where the curves meet and help define the interval over which you will integrate. By setting the functions equal to each other and solving for $x$ , you can find these intersection points and use them to accurately calculate the area between the curves.

Created using AI

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Area Between Curves: Videos & Practice Problems

Finding Area Between Curves on a Given Interval

Video transcript

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

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

Shade the region bounded by f(x)=log2x+4\textcolor{blue}{f\left(x\right)=log_2x+4} & g(x)=ex−4\textcolor{orange}{g\left(x\right)=e^{x-4}} on the interval [1,4]. Then set up an integral to represent the region's area.

Finding Area Between Curves on a Given Interval

Video transcript

Finding Area When Bounds Are Not Given

Video transcript

Find the area between f(x)=x2−4f\left(x\right)=x^2-4f(x)=x2−4 & g(x)=−x2+4g\left(x\right)=-x^2+4g(x)=−x2+4.

Find the area of the shaded region between f(x)=x4−x2f\left(x\right)=x^4-x^2f(x)=x4−x2 & g(x)=3x2g\left(x\right)=3x^2g(x)=3x2.

Find the area of the shaded region shown in the first quadrant between f(x)=1xf\left(x\right)=\frac{1}{x} & g(x)=−14x+54.g\left(x\right)=-\frac14x+\frac54.

Finding Area Between Curves that Cross on the Interval

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Finding Area Between Curves that Cross on the Interval Example 2

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Find the shaded area between f(x)=x3+2x2\textcolor{orange}{f\left(x\right)=x^3+2x^2}f(x)=x3+2x2 & g(x)=x+2\textcolor{blue}{g\left(x\right)=x+2}g(x)=x+2.

Find the area of the shaded region between f(x)=1xf\left(x\right)=\frac{1}{x} & g(x)=xg\left(x\right)=x from x=0.5x=0.5 to x=4x=4.

Finding Area Between Curves that Cross on the Interval Example 3

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Here’s what students ask on this topic:

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

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

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

Can you find the area between curves if the functions are below the x-axis?

What is the significance of the intersection points when finding the area between curves?

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

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

Shade the region bounded by $f (x) = l o g_{2} x + 4$ & $g (x) = e^{x - 4}$ on the interval [1,4]. Then set up an integral to represent the region's area.

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

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

Find the area of the shaded region shown in the first quadrant between $f (x) = \frac{1}{x}$ & $g (x) = - \frac{1}{4} x + \frac{5}{4} .$

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

Find the area of the shaded region between $f (x) = \frac{1}{x}$ & $g (x) = x$ from $x equals 0.5$ to $x equals 4$ .