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This page covers double integrals in rectangular coordinates. Double integrals in polar coordinates are covered on a separate page.
Alternate Name For Double Integrals - Iterated Integrals |
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Double integrals are just integrals that are nested, i.e. there are two integrals, one inside of the other. The idea is to evaluate each integral separately, starting with the inside integral.
The discussion on this page is in two main parts based on the type of region described by the limits of integration.
If the limits of integration are constant, the region is rectangular. |
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If the limits of integration are NOT constant, the region is non-rectangular. |
The basics of double integrals are discussed first in the context of rectangular regions. Then alternate techniques are discussed to handle non-rectangular regions. Before we get started on the details of evaluating double integrals, let's watch a video to help understand what double integrals represent.
video by Krista King Math
Rectangular Regions |
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Rectangular regions in the x-y plane are indicated by constants in the limits of integration, i.e. \( a \leq x \leq b \) and \( c \leq y \leq d\).
\(\displaystyle{ \int_{c}^{d}{ \int_{a}^{b}{g(x,y) ~ dx} ~dy } = \int_{c}^{d}{ \left[ \int_{a}^{b}{g(x,y) ~ dx}\right] ~dy}}\)
You can think about it this way.
First, evaluate the inside integral with respect to the inside variable, while holding the other variable constant. In the integral above, the inside variable is x.
Second, take the result of the first integration and now integrate with respect to the outside variable. In the above integral, the outside variable is y.
This integral, once evaluated, gives the area or volume above the x-y plane with height \(g(x,y)\). We have more discussion of area and volume with double integrals on the area and volume application page. For now, we will just discuss how to evaluate a double integral.
Notice the order \(dx~dy\) or \(dy~dx\) is critical in determining which limits of integration go with which variable. In the example above, we integrated on the interval \([a,b]\) in the x direction and we evaluated on the interval \([c,d]\) in the y direction.
Okay, so that is a very basic introduction on how to look at a double integral. Let\'s watch a video to get started. This video explains the motivation behind double integrals, explains what double integrals are and includes the basics on how to set them up and evaluate them, using some basic examples.
video by Dr Chris Tisdell
Are you ready for an example? Let's look at an easy double integral and take it step-by-step.
Evaluate \(\displaystyle{ \int_{1}^{2}{ \int_{2}^{4}{3xy^2 ~ dx} ~dy } }\)
42
\(\displaystyle{ \int_{1}^{2}{ \int_{2}^{4}{3xy^2 ~ dx} ~dy } }\) |
\(\displaystyle{ \int_{1}^{2}{ \int_{2}^{4}{3xy^2 ~ dx} ~dy } }\) |
\(\displaystyle{ \int_{1}^{2}{ \left[ y^2 \int_{2}^{4}{3x ~ dx} \right] ~dy } }\) |
\(\displaystyle{ \int_{1}^{2}{ y^2 \left[ 3x^2/2 \right]_{x=2}^{x=4} ~dy } }\) |
\(\displaystyle{ \int_{1}^{2}{ y^2 \left[ 3(4)^2/2 - 3(2)^2/2 \right] ~dy } }\) |
\(\displaystyle{ \int_{1}^{2}{ y^2 \left[ 18 \right] ~dy } = \int_{1}^{2}{ 18y^2 ~dy } }\) |
\( \left[ 18y^3/3 \right]_{1}^{2} \) |
\( 18(2)^3/3 - 18(1)^3/3 \) |
\( 48 - 6 = 42 \) |
Notice in this solution, that after evaluating the inside integral, all the x's are gone, so that the outside integral is just a basic single variable definite integral with integrand \(18y^2\). This is due to the fact that we have constants as the limits of integration for the inside integral. Makes sense, right?
Final Answer |
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42 |
close solution |
Okay, let's watch a video tutorial. Here is another very good video explaining double integrals and how they work.
video by Dr Chris Tisdell
In the above example, we were able to use a special technique that works only in specific instances. Here is a great video that explains exactly when and how to use it. Although we do not usually encourage learning special cases, this is easy to use, works all the time ( when used correctly ), saves time and it is easy to learn. So go for it!
video by Dr Chris Tisdell
Notation |
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As discussed in the videos above, a double integral can be thought of as an integral over a region in the plane. This can be written several different ways. If you are given a region in the plane defined by the region R, all of the forms below describe the same integral.
the integral of \(f(x,y)\) over the region \( R = \{ (x,y) ~:~ a \leq x \leq b, ~ c \leq y \leq d \} \) | ||
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\(\displaystyle{ \int_{c}^{d}{ \int_{a}^{b}{f(x,y) ~ dx} ~dy } }\) |
\(\displaystyle{ \iint\limits_R {f(x,y) ~ dA} }\) |
In the last form, \(dx~dy\) are combined to form \(dA\) and the integral signs are collapsed into the integral over R. This notation is very common and it is something you will see more of as you continue on in calculus.
Reversing Order of Integration (Rectangular Regions) |
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When the limits of integration of both integrals are constants, we can switch the limits of integration with no other changes to the integral, without affecting the result. Stop and think about this. This holds only when the limits of integration are constants.
So in the example above, we could have integrated with respect to y first, then with respect to x and arrived at the same answer. The equivalent integrals are
\(\displaystyle{ \int_{1}^{2}{ \int_{2}^{4}{3xy^2 ~ dx} ~dy } = \int_{2}^{4}{ \int_{1}^{2}{3xy^2 ~ dy} ~dx } }\)
Notice that when we switched the integral signs on the left, we also switched \(dx~dy \to dy~dx\). This is required since the intervals do not change, i.e. \([2,4]\) is in the x direction and \([1,2]\) is in the y direction. So we need to keep the same association in an equivalent integral. [ Try working this second integral to convince yourself of this. ]
Before we look at non-rectangular regions, let's work some practice problems. Unless otherwise instructed, evaluate the following integrals, giving your answers in exact terms. If it is not possible to evaluate the integral in the given form, try changing the order of integration.
\(\displaystyle{ \int_{1}^{3}{ \int_{2}^{3}{ x^2 - 2xy + 2y^3 ~ dy } ~dx } }\)
Problem Statement |
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\(\displaystyle{ \int_{1}^{3}{ \int_{2}^{3}{ x^2 - 2xy + 2y^3 ~ dy } ~dx } }\)
Solution |
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video by Dr Chris Tisdell
close solution |
\(\displaystyle{ \int_{2}^{3}{ \int_{0}^{1}{ x^2 y ~dx } ~dy } }\)
Problem Statement |
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\(\displaystyle{ \int_{2}^{3}{ \int_{0}^{1}{ x^2 y ~dx } ~dy } }\)
Solution |
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video by Dr Chris Tisdell
close solution |
\(\displaystyle{ \int_{1}^{3}{ \int_{0}^{2}{ xy + x^2y^3 ~dy } ~dx } }\)
Problem Statement |
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\(\displaystyle{ \int_{1}^{3}{ \int_{0}^{2}{ xy + x^2y^3 ~dy } ~dx } }\)
Solution |
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video by PatrickJMT
close solution |
Integrate \(\displaystyle{ \frac{2y}{(x^2+1)} }\) over the region \( \{ (x,y) ~|~ 0 \leq x \leq 1,~0 \leq y \leq \sqrt{x} \} \)
Problem Statement |
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Integrate \(\displaystyle{ \frac{2y}{(x^2+1)} }\) over the region \( \{ (x,y) ~|~ 0 \leq x \leq 1,~0 \leq y \leq \sqrt{x} \} \)
Solution |
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video by PatrickJMT
close solution |
\(\displaystyle{ \int_{0}^{2}{ \int_{0}^{4}{ 3x + 4y ~dx } ~dy } }\)
Problem Statement |
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\(\displaystyle{ \int_{0}^{2}{ \int_{0}^{4}{ 3x + 4y ~dx } ~dy } }\)
Solution |
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video by Krista King Math
close solution |
\(\displaystyle{ \int_{-1}^{2}{ \int_{1}^{3}{ 2x - 7y ~dy } ~dx } }\)
Problem Statement |
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\(\displaystyle{ \int_{-1}^{2}{ \int_{1}^{3}{ 2x - 7y ~dy } ~dx } }\)
Solution |
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video by Krista King Math
close solution |
\(\displaystyle{ \int_{0}^{3}{ \int_{0}^{3}{ xy + 7x + y ~dx } ~dy } }\)
Problem Statement |
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\(\displaystyle{ \int_{0}^{3}{ \int_{0}^{3}{ xy + 7x + y ~dx } ~dy } }\)
Solution |
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video by Krista King Math
close solution |
Integrate \(\displaystyle{ f(x,y) = 2xy - 3y^2 }\) over the region \( R=[-1,1] \times [-2,2] \)
Problem Statement |
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Integrate \(\displaystyle{ f(x,y) = 2xy - 3y^2 }\) over the region \( R=[-1,1] \times [-2,2] \)
Solution |
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video by Krista King Math
close solution |
\(\displaystyle{ \int_{1}^{4}{ \int_{1}^{2}{ \frac{x}{y} + \frac{y}{x} dy} ~dx } }\)
Problem Statement |
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\(\displaystyle{ \int_{1}^{4}{ \int_{1}^{2}{ \frac{x}{y} + \frac{y}{x} dy} ~dx } }\)
Solution |
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video by Krista King Math
close solution |
\(\displaystyle{ \iint\limits_R{ \frac{xy^2}{x^2+1} ~dA } }\) \( R=\{(x,y)~:~ 0 \leq x \leq 1, ~ -3 \leq y \leq 3 \} \)
Problem Statement |
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\(\displaystyle{ \iint\limits_R{ \frac{xy^2}{x^2+1} ~dA } }\) \( R=\{(x,y)~:~ 0 \leq x \leq 1, ~ -3 \leq y \leq 3 \} \)
Solution |
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video by Krista King Math
close solution |
\(\displaystyle{ \int_{0}^{ \sqrt{2\pi/3}}{ \int_{0}^{\sqrt{7\pi/6}}{ xy ~ \sin(x^2+y^2) ~dx } ~dy } }\) |
Problem Statement |
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\(\displaystyle{ \int_{0}^{ \sqrt{2\pi/3}}{ \int_{0}^{\sqrt{7\pi/6}}{ xy ~ \sin(x^2+y^2) ~dx } ~dy } }\) |
Final Answer |
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\( \sqrt{3}/8 \) |
Problem Statement |
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\(\displaystyle{ \int_{0}^{ \sqrt{2\pi/3}}{ \int_{0}^{\sqrt{7\pi/6}}{ xy ~ \sin(x^2+y^2) ~dx } ~dy } }\) |
Solution |
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video by MIP4U
Final Answer |
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\( \sqrt{3}/8 \) |
close solution |
\(\displaystyle{ \int_{1/2}^{1}{ \int_0^1{ \frac{y^2-x^2}{(x^2+y^2)^2} ~dx } ~dy } }\)
Problem Statement |
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\(\displaystyle{ \int_{1/2}^{1}{ \int_0^1{ \frac{y^2-x^2}{(x^2+y^2)^2} ~dx } ~dy } }\)
Hint |
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Use the trig substitution \( x = y\tan\theta \).
Problem Statement |
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\(\displaystyle{ \int_{1/2}^{1}{ \int_0^1{ \frac{y^2-x^2}{(x^2+y^2)^2} ~dx } ~dy } }\)
Final Answer |
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\(\pi/4 - \arctan(1/2)\) |
Problem Statement |
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\(\displaystyle{ \int_{1/2}^{1}{ \int_0^1{ \frac{y^2-x^2}{(x^2+y^2)^2} ~dx } ~dy } }\)
Hint |
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Use the trig substitution \( x = y\tan\theta \).
Solution |
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video by PatrickJMT
Final Answer |
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\(\pi/4 - \arctan(1/2)\) |
close solution |
Non-Rectangular Regions |
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So, now we add a level of complexity where the region in the x-y plane is not rectangular. This results in the limits of integration not being constant. There are only a couple of additional things that need to happen when setting up and evaluating double integrals with non-constant limits.
First - - It is imperative to plot the region in the plane when setting up the integral. |
Second - - When switching the limits of integration, the limits themselves will change. |
Plotting The Region |
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As mentioned above, the double integral can be thought of as representing the volume of a solid above a region in the x-y plane of height \(f(x,y)\) (the integrand). When working with non-rectangular regions, you need to plot the x-y region. This will help you determine the limits of integration. Let us make this clear. This is not something that is optional that we are suggesting. This is a critical step in solving these problems and we have never found anyone that can do without it by doing it in their head. In fact, most instructors will require a plot for full credit on homework and exams. So don't neglect doing this in order to save time or energy or whatever. It's not worth it.
To help a bit and if your instructor allows it, you can use the free program winplot to plot the boundaries and then print it out and work with it for your homework. A link to the free program is on the tools page. We use winplot for many graphs on this site. But check with your instructor to make sure it is okay with them. This next panel explains how to describe the region once you have graphed it.
To describe an area in the xy-plane, the first step is to plot the boundaries and determine the actual region that needs to be described. There are several graphing utilities listed on the tools page. Our preference is to use the free program winplot ( used to plot these graphs; we used gimp to add labels and other graphics ). However, graphing by hand is usually the best and quickest way.
We use the graph to the right to facilitate this discussion. A common way to describe this area is the area bounded by \(f(x)\) (red line), \(g(x)\) (blue line) and \(x=a\) (black line).
[Remember that an equation like \(x=a\) can be interpreted two ways, either the point x whose value is a or the vertical line. You should be able to tell what is meant by the context.]
Okay, so we plotted the boundaries and shaded the area to be described. Now, we need to choose a direction to start, either vertically or horizontally. We will show both ways, starting with vertically, since it is more natural and what you are probably used to seeing. Also, this area is easier to describe vertically than horizontally (you will see why as you read on).
Vertically |
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Our first step is to draw a vertical arrow on the graph somewhere within the shaded area, like we have done here. Some books draw an example rectangle with the top on the upper graph and the bottom on the lower graph. That is the same idea as we have done with the arrow.
Now we need to think of this arrow as starting at the left boundary and sweeping across to the right boundary of the area. This sweeping action is important since it will sweep out the area. As we think about this sweeping, we need to think about where the arrow enters and leaves the shaded area. Let's look our example graph to demonstrate. Think about the arrow sweeping left to right. Notice that it always enters the area by crossing \(g(x)\), no matter where we draw it. Similarly, the arrow always exits the area by crossing \(f(x)\), no matter where we draw it. Do you see that?
But wait, how far to the right does it go? We are not given that information. What we need to do is find the x-value where the functions \(f(x)\) and \(g(x)\) intersect. You should be able to do that. We will call that point \((b,f(b))\). Also, we will call the left boundary \(x=a\). So now we have everything we need to describe this area. We give the final results below.
Vertical Arrow | |
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\( g(x) \leq y \leq f(x) \) |
arrow leaves through \(f(x)\) and enters through \(g(x)\) |
\( a \leq x \leq b \) |
arrow sweeps from left (\(x=a\)) to right (\(x=b\)) |
Horizontally |
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We can also describe this area horizontally (or using a horizontal arrow). We will assume that we can write the equations of \(f(x)\) and \(g(x)\) in terms of \(y\). ( This is not always possible, in which case we cannot describe the area in this way. ) For the sake of this discussion, we will call the corresponding equations \(f(x) \to F(y)\) and \(g(x) \to G(y)\).
Let's look at the graph. Notice we have drawn a horizontal arrow. Just like we did with the vertical arrow, we need to determine where the arrow enters and leaves the shaded area. In this case, the arrow sweeps from the bottom up. As it sweeps, we can see that it always crosses the vertical line \(x=a\). However, there is something strange going on at the point \((b,f(b))\). Notice that when the arrow is below \(f(b)\), the arrow exits through \(g(x)\) but when the arrow is above \(f(b)\), the arrow exits through \(f(x)\). This is a problem. To overcome this, we need to break the area into two parts at \(f(b)\).
Lower Section - - This section is described by the arrow leaving through \(g(x)\). So the arrow sweeps from \(g(a)\) to \(g(b)\).
Upper Section - - This section is described by the arrow leaving through \(f(x)\). The arrow sweeps from \(f(b)\) to \(f(a)\).
The total area is the combination of these two areas. The results are summarized below.
Horizontal Arrow | |
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lower section | |
\( a \leq x \leq G(y) \) |
arrow leaves through \(G(y)\) and enters through \(x=a\) |
\( g(a) \leq y \leq g(b) \) |
arrow sweeps from bottom (\(y=g(a)\)) to top (\(y=g(b)\)) |
upper section | |
\( a \leq x \leq F(y) \) |
arrow leaves through \(F(y)\) and enters through \(x=a\) |
\( f(b) \leq y \leq f(a) \) |
arrow sweeps from bottom (\(y=f(b)\)) to top (\(y=f(a)\)) |
Type 1 and Type 2 Regions |
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Some instructors may describe regions in the plane as either Type 1 or Type 2 (you may see II instead of 2). As you know from the above discussion, some regions are better described vertically or horizontally. Type 1 regions are regions that are better described vertically, while Type 2 regions are better described horizontally. The example above was a Type 1 region.
Here is a quick video clip going into more detail on Type 1 and Type 2 regions.
video by Krista King Math
[ For some videos and practice problems dedicated to this topic, check out this page. ]
Reversing Order of Integration (Non-Rectangular Regions) |
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So what do we do when the limits of integration are NOT constants and why would we want to switch the limits of integration anyway?
Let's answer the second question first. This is certainly a valid question since we don't just want to add extra work if we don't need to. And that's the key . . . do we really need to? The answer is yes. Sometimes the inside integral can not be evaluated, which means that the outside integral cannot be evaluated either. So, switching the order of integration will often allow us to evaluate the inside integral. You will see why soon.
When the area is not rectangular, the procedure is completely different for reversing the order of integration than we saw when the area was rectangular. The first critical step is to plot the region over which the integral is being evaluated. After plotting, you need to describe the region differently. It is time for a video to help you understand what you need to do.
Here is a great video that demonstrates, in general, how to switch the limits of integration for a non-rectangular region. Notice that he plots the region and how critical the plot is to his solution.
video by PatrickJMT
A couple of things that he showed in that last video require comments.
First - - Labeling the limits of integration showing which variable it was assigned to, is a powerful technique to keep track of the limits. Try to get in the habit of doing this ( if your instructor allows it, of course ), since later on ( with triple integrals ), it can help you keep track of what is going on.
Second - - He drew an arrow to represent the limits of the inside limits which is extremely helpful. However, to represent the outside limits, it is better to think of the outside integral as sweeping that arrow across the region, which seems to sweep out the area itself.
Here is a video clip explaining some properties of double integrals.
video by Dr Chris Tisdell
Okay, so now you have the basics. Time for some practice problems. After that, the next page you may want to go to explains using double integrals to calculate areas and volumes in more detail.
Instructions - - Unless otherwise instructed, evaluate the following integrals, giving your answers in exact terms. If it is not possible to evaluate the integral in the given form, try changing the order of integration.
Integrate \( f(x,y) = x^2y+y^3 \) over the region \( \{(x,y) ~ \left| ~ \right. x^2 + y^2 \leq 1, ~ x \geq 0, ~ y \geq 0 \} \)
Problem Statement |
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Integrate \( f(x,y) = x^2y+y^3 \) over the region \( \{(x,y) ~ \left| ~ \right. x^2 + y^2 \leq 1, ~ x \geq 0, ~ y \geq 0 \} \)
Solution |
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video by Dr Chris Tisdell
close solution |
Change the order of integration and evaluate \(\displaystyle{ \int_{0}^{1}{ \int_{y}^{1}{ e^{x^2} dx } ~dy } }\)
Problem Statement |
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Change the order of integration and evaluate \(\displaystyle{ \int_{0}^{1}{ \int_{y}^{1}{ e^{x^2} dx } ~dy } }\)
Solution |
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video by Dr Chris Tisdell
close solution |
Change the order of integration and evaluate \(\displaystyle{ \int_{0}^{1}{ \int_{x}^{1}{ \left( 1 - y^2 \right)^{-1/2} dy } ~dx } }\)
Problem Statement |
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Change the order of integration and evaluate \(\displaystyle{ \int_{0}^{1}{ \int_{x}^{1}{ \left( 1 - y^2 \right)^{-1/2} dy } ~dx } }\)
Solution |
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video by Dr Chris Tisdell
close solution |
\(\displaystyle{ \int_{0}^{1}{ \int_{ \sqrt{x}}^{1}{ \sqrt{1+y^3} ~dy } ~dx } }\)
Problem Statement |
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\(\displaystyle{ \int_{0}^{1}{ \int_{ \sqrt{x}}^{1}{ \sqrt{1+y^3} ~dy } ~dx } }\)
Solution |
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video by Dr Chris Tisdell
close solution |
\(\displaystyle{ \int_{0}^{1}{ \int_{y}^{1}{ \sin(x^2) ~dx } ~dy } }\)
Problem Statement |
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\(\displaystyle{ \int_{0}^{1}{ \int_{y}^{1}{ \sin(x^2) ~dx } ~dy } }\)
Solution |
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video by Dr Chris Tisdell
close solution |
\(\displaystyle{ \int_{0}^{1}{ \int_{y^2}^{1}{ 2\sqrt{x}~e^{x^2} ~dx } ~dy } }\)
Problem Statement |
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\(\displaystyle{ \int_{0}^{1}{ \int_{y^2}^{1}{ 2\sqrt{x}~e^{x^2} ~dx } ~dy } }\)
Solution |
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video by Dr Chris Tisdell
close solution |
\(\displaystyle{ \int_{0}^{2}{ \int_{x^2}^{4}{ \frac{x^3}{\sqrt{x^4+y^2}} ~dy } ~dx } }\)
Problem Statement |
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\(\displaystyle{ \int_{0}^{2}{ \int_{x^2}^{4}{ \frac{x^3}{\sqrt{x^4+y^2}} ~dy } ~dx } }\)
Solution |
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video by Dr Chris Tisdell
close solution |
Evaluate \(\displaystyle{ \int_{-2\pi}^0{ \int_{-y^2}^{\cos(y^2)}{ 2y ~dx } ~dy } }\)
Problem Statement |
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Evaluate \(\displaystyle{ \int_{-2\pi}^0{ \int_{-y^2}^{\cos(y^2)}{ 2y ~dx } ~dy } }\)
Final Answer |
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\( -\sin(4\pi^2) - 8\pi^4 \) |
Problem Statement |
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Evaluate \(\displaystyle{ \int_{-2\pi}^0{ \int_{-y^2}^{\cos(y^2)}{ 2y ~dx } ~dy } }\)
Solution |
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video by World Wide Center of Mathematics
Final Answer |
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\( -\sin(4\pi^2) - 8\pi^4 \) |
close solution |