Given a dual integraleginalign* iint_dlr f(x,y),dAendalign*of a function $f(x,y)$ end a an ar $dlr$, friend may be able to write itas two various iterated integrals. Girlfriend can integrate withrespect come $x$ first, or girlfriend can incorporate with respect to $y$ first.If you incorporate with respect to $x$ first, you will achieve anintegral the looks something likeeginalign* iint_dlr f(x,y),dA = int_Box^Box left(int_Box^Box f(x,y),dx
ight) dy,endalign*and if you incorporate with respect to $y$ first, friend will achieve anintegral that looks something likeeginalign* iint_dlr f(x,y),dA = int_Box^Box left(int_Box^Box f(x,y),dy
ight) dx.endalign*We regularly say the the an initial integral is in $dx,dy$ order and the second integral is in $dy,dx$ order.

You are watching: Evaluate the integral by reversing the order of integration

One an overwhelming part that computing double integrals is determining the limits of integration, i.e., identify what to put in ar of the boxes $Box$ in the above integrals. In some situations, we understand the borders of integration the $dx,dy$ order and also need to recognize the limits of integration for the identical integral in $dy,dx$ order (or evil versa). The process of switching in between $dx,dy$ order and also $dy,dx$ stimulate in twin integrals is called transforming the bespeak of integration (or reversing the order of integration).

Changing the bespeak of integration is slightly tricky since its hardto create down a specific algorithm because that the procedure. The easiest wayto achieve the task is through illustration a photo of the region$dlr$. From the picture, you deserve to determine the corners and edges the theregion $dlr$, i m sorry is what you need to write down the boundaries ofintegration.

We show this procedure with examples. The simplest region (other than a rectangle) for reversing the integration bespeak is a triangle. You have the right to see just how to change the stimulate of integration for a triangle by comparing example 2 with instance 2" on the web page of twin integral examples. In this page, we offer some more examples an altering the integration order.

Example 1Change the order of integration in the adhering to integral eginalign* int_0^1 int_1^e^y f(x,y) dx, dy.endalign*(Since the focus of this example is the limits of integration, us won"t clues the role $f(x,y)$. The procedure doesn"t depend on the identification of $f$.)

**Solution**: In the original integral, the integration stimulate is $dx,dy$. This integration order synchronizes to integrating first with respect come $x$ (i.e., summing along rows in the snapshot below), and after that integrating with respect to $y$ (i.e., summing increase the values for each row). Our task is to readjust the integration to it is in $dy,dx$, which way integrating an initial with respect to $y$.

We begin by transforming the limits of integration right into the domain $dlr$. The limits of the outer $dy$ integral average that$0 le y le 1,$and the borders on the inside $dx$ integral typical that for each worth of $y$ the range of $x$ is$1 le x le e^y.$The an ar $dlr$ is shown in the complying with figure.

The maximum variety of $y$ over the region is native 0 to 1, as suggested by the gray bar come the left of the figure. The horizontal hashing in ~ the number indicates the variety of $x$ for each worth of $y$, beginning at the left edge $x=1$ (blue line) and also ending at the best curve edge $x=e^y$ (red curve).

We have likewise labeled all the corners the the region. The upper-right corner is the intersection of the heat $y=1$ v the curve $x=e^y$. Therefore, the worth of $x$ at this edge must it is in $e=e^1=e$, and the allude is $(e,1)$.

To change order that integration, we have to write one integral with order $dy,dx$. This method that $x$ is the change of the external integral. Its borders must be constant and correspond to the total range of $x$ end the an ar $dlr$. The total variety of $x$ is $1 le x le e$, as indicated by the gray bar below the an ar in the adhering to figure.

Since $y$ will certainly be the variable because that the inside integration, we require to combine with respect to $y$ first. The upright hashing indicates how, because that each value of $x$, we will integrate from the reduced boundary (red curve) come the top boundary (purple line). These two limits determine the selection of $y$. Since we have the right to rewrite the equation $x=e^y$ because that the red curve as $y=log x$, the selection of $y$ is $log x le y le 1$. (The duty $log x$ suggests the natural logarithm, which occasionally we create as $ln x$.)

In summary, the an ar $dlr$ can be explained not only by egingather* 0 le y le 1\ 1 le x le e^yendgather*as it was for the initial $dx,dy$ integral, but also byegingather* 1 le x le e\ log x le y le 1,endgather*which is the summary we require for the new $dy,dx$ integration order. This latter pair that inequalites recognize the bounds because that integral.

We conclude the the integral$int_0^1 int_1^e^y f(x,y) dx, dy$ with integration order reversed iseginalign* int_1^e int_log x^1 f(x,y) dy , dx.endalign*

Example 2Sometimes you need to change the order of integration to get atractable integral. Because that example, if girlfriend tried come evaluateeginalign* int_0^1 int_x^1 e^y^2dy,dxendalign*directly, you would run into trouble. Over there is no antiderivative of$e^y^2$, for this reason you get stuck trying to compute the integral withrespect to $y$. But, if we change the stimulate of integration, climate wecan combine with respect to $x$ first, which is doable. And, itturns the end that the integral through respect come $y$ also becomes possibleafter we finish integrating through respect to $x$.

According to the borders of integration the the provided integral, the an ar of integration isegingather* 0 le x le 1\ x le y le 1,endgather*which is presented in the adhering to picture.

Since us can additionally describe the region by egingather* 0 le y le 1\ 0 le x le y,endgather*the integral through the order adjusted iseginalign* int_0^1 int_x^1 e^y^2 dy,dx = int_0^1int_0^y e^y^2 dx,dyendalign*With this brand-new $dx,dy$ order, we integrate an initial with respect come $x$eginalign*int_0^1int_0^y e^y^2 dx,dy = int_0^1 x left.left.e^y^2 ight|_x=0^x=y ight. Dy = int_0^1 y e^y^2 dy.endalign*Since the integration with respect to $x$ provided us an extra element of $y$, we can compute the integral with respect to $y$ by making use of a$u$-substitution, $u=y^2$, so $du=2y,dy$. With this substitution, $u$ rannges indigenous 0to 1, and we calculation the integral aseginalign*int_0^1int_0^y e^y^2 dx,dy &=int_0^1 y e^y^2 dy\ &= int_0^1frac12 e^u du = frac12 e^u igg|_0^1 = frac12(e-1).endalign*

Example 3Here"s an instance that"s a bit more tricky. Reverse the bespeak of integration in the adhering to integral.egingather* int_pi/2^5pi/2 int_sin x^1 f(x,y) dy,dxendgather*

**Solution**: The region $dlr$ explained by this integral isegingather* pi/2 le x le 5pi/2\ sin x le y le 1.endgather*as displayed in the following image, whereby the total range on $x$ is presented by the gray bar listed below the region, and also the variable boundaries for $y$ are presented by the blue and cyan curves.

One cheat for an altering variables through this an ar is correctly managing the lower boundary $y = sin(x)$. Once we deal with this border equation for $x$ together a role of $y$, we might be tempted to write it as $x = arcsin(y)$ and also maybe also think the $x le arcsin(y)$ in the region.

Looking carefully at the picture, we see this can not be the case. In fact, the reduced boundary because that $y$ together a duty of $x$ (the blue curve) needs to be both the upper and also lower borders for $x$ together a role of $y$, as presented by the red and also purple curve in the below figure.

To gain the formula because that these boundaries, we need to remember how the station of the sinusoid, $arcsin(y)$, is defined. In bespeak to define the station of $sin(x)$, we need to restrict the function to an interval wherein it bring away on every value just one time. The standard means to specify $arcsin(y)$ is to restrict $sin(x)$ to worths of $x$ in the term $<-pi/2,pi/2>$ together $sin(x)$ ranges from $-1$ to 1 in the interval. This way that $arcsin(y)$ varieties from $<-pi/2,pi/2>$ as $y$ goes native $-1$ to 1.

For the upper boundary of $x$ (in purple), $x$ ranges from $3pi/2$ come $5pi/2$. If us let $x=arcsin(y)+2pi$, then $x=3pi/2$ once $y=-1$ and $x=5pi/2$ when $y=1$, together required. Because that the lower boundary that $x$ (in red), we need $x$ to be a decreasing function of $y$, starting at $x=3pi/2$ as soon as $y=-1$ and decreasing to $x=pi/2$ as soon as $y=1$. These problems are satisfied if we choose $x=pi-arcsin(y)$. If you space an professional at your trignometric identifies, you can verify that the equations for both of this curves room just various inverses of $sin(x)$, as taking the sinusoid of this equations reduce them to $y=sin(x)$.

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Since in the region, $y$ ranges over the interval $<-1,1>$ (gray bar come the left of the region), us can describe the an ar $dlr$ through the inequalitiesegingather* -1 le y le 1\ pi-arcsin y le x le arcsin y+2pi.endgather*This description of $dlr$ is what we require to adjust the order of integration, and we find thategingather* int_pi/2^5pi/2 int_sin x^1 f(x,y) dy,dx = int_-1^1 int_pi-arcsin y^arcsin y+2pi f(x,y) dx,dy.endgather*

More examplesIf you"d prefer more twin integral examples, you have the right to study part introductorydouble integral examples.. Friend can additionally take a look at at double integral examples from the special situations of interpreting dual integrals as area and double integrals together volume.