how to reverse order of integration
Video How to reverse the order of integrals Provide a double integral begin {align *} iint_dlr f(x, y), dA end {align *} of the function $f(x,y)$ over a region $dlr$, you it might be possible to write it as two different repeating integrals. You can integrate with $ x $ first or you can integrate with $ y $ first. If you integrate with respect to $x $ first, you will get an integral that looks like begin {align *} iint_dlr f(x, y), dA = int_ {Box} ^ {Box} left (int_ {Box} ^ {Box} f (x, y), dx right) dy, end {align *} and if you integrate with respect to $y $ first, you get an integral that looks like begin {align *} iint_dlr f (x , y), dA = int_ {Box} ^ {Box} left (int_ {Box} ^ {Box} f (x, y), dy right) dx. end {align *} We often say that the first integral is in the order $ dx, dy $ and the second integral is in the order $ dy, dx $. One difficult part of calculating double integrals is determining the boundary. term of the integral, i.e. determine what to put in the place of the boxes $Box$ in the above integral. In some situations, we know the limit of integrals of the order $dx,dy$ and need to define the limit of integration for the equivalent integral of the order $dy,dx$ (or vice versa). The transition between $dx, dy $order and $dy,dx $order in a double integral is called changing the order of integrals (or reversing the order of integrals). It’s a bit complicated because it’s hard to write a process-specific algorithm. The easiest way to accomplish the task is to draw a picture of the $dlr$ area. From the image you can determine the corners and edges of the $dlr$ region, which is what you need to write out the bounds of the integration. We demonstrate this process with examples. The simplest region (not a rectangle) to reverse the integration order is a triangle. You can see how to change the order of integration for a triangle by comparing example 2 with example 2 ′ on the double integral examples page. In this page we give some more examples of changing the order of integrals Example 1 Changing the order of integrals in the integral after begin {align *} int_0^1 int_1^{e^y} f ( x, y) dx, dy. end {align *} (Since the focus of this example is on the limitations of integration, we will not specify the function $f(x, y)$. This procedure does not depend on the identity of $f$. .) Solution: In the original integral, the order of integration is $dx,dy$. This order of integration corresponds to first integrating for $x$ (i.e. summing along the rows in the figure below) and then integrating for $y$ (i.e. summing the values for per row). Our task is to change integration to $dy,dx$, which means integrate first for $y$. We start by converting the integration limits to the $dlr$ domain. The limit of the outer $dy $ integral means $0 le y le 1, and the limit of the inner $ dx $ integral means that for each value of $ y $, the range of $ x $ is $1 le x le e ^ y. $ The $dlr $ zone is shown in the following figure.The maximum range of $y$ in the region is from 0 to 1, as indicated by the gray bar to the left of the figure. The horizontal hash in the figure represents the range $x$ for each value of $y$, starting at the left edge $x=1$ (blue line) and ending at the right edge of the curve $x=e ^y$ (red curve). Read more: how to draw an owl easy We have also labeled all the corners of the area. The upper right corner is the intersection of the line $y=1$ with the curve $x=e^y$. Therefore, the value of $x$ at this angle should be $e = e^1 = e$, and the point is $(e, 1)$. To change the order of integrals, we need to write an integral of the order $dy,dx$. This means that $ x $ is the variable of the outer integral. Its limit must be constant and corresponds to the total range $x$ over the $dlr$ region. The total range of $ x $ is $ 1 le x le e $, as indicated by the gray bar below the area in the following figure.
Since $y$ will be the variable for internal integration, we need to integrate with $y$ first. The vertical hash shows how, for each $ x $ value, we will integrate from the lower boundary (red curve) to the upper boundary (purple line). These two boundaries define the range of $y$. Since we can rewrite the equation x = e^y$ for the red curve as $y = log x $, the range of $y $ is $ log x le y le 1 $. (The $log x $ function represents the natural logarithm, which we sometimes write as $ln x $.) In short, the $dlr$ region can be described more than just begin {collect *} 0 le y le 1 1 le x le e ^ y end {collect *} as for the integral $dx, dy $ initially, but also because begin {collect *} 1 le x le e log x le y le 1, end {collect *} which is the description we need for the new $dy,dx$ integration order. This second pair of inequalities defines the limits for integration We conclude that the integral $int_0^1 int_1^{e^y}f(x,y)dx,dy$ with the order of integration reversed the reverse is begin {align *} int_1 ^ e int_ {log x} ^ 1 f (x, y) dy, dx. end {align *} Example 2 Sometimes you need to change the order of integrals to get an adjustable integral. For example, if you try to evaluate begin {align *} int_0^1 directly int_x^1 e ^ {y ^ 2} dy, dx end {align *} directly, you’ll run into trouble. There is no differential of e^{y^2}$, so you have a hard time trying to integrate with respect to y$. However, if we change the order of integration, then we can integrate for $x$ first, which is doable. And, it turns out that the integration with respect to y $ can also become possible after we finish the integration with respect to $ x $. According to the integral limit of the given integral, the integral region is started {collect *} 0 le x le 1 x le y le 1, end {collect *} is shown in the following figure.
Since we can also describe the region by begin {collect *} 0 le y le 1 le x le y, end {collect *} the integral has the changed order as begin {align *} int_0^1 int_x^1 e^{y^2} dy, dx = int_0^1int_0^ye^{y^2} dx,dy end {align *} With this new $dx,dy$ order, we pre-integrate with $ x $ begin { align * } int_0 ^ 1int_0 ^ ye ^ {y ^ 2} dx, dy = int_0 ^ 1 x left.left.e ^ {y ^ 2} right | _ {x = 0} ^ {x = y} right. dy = int_0 ^ 1 ye ^ {y ^ 2} dy. end {align *} Since the integral over $x$ gave us an extra factor of $y$, we can integrate with respect to $y$ using $u $-sub, $u = y^2$, so $du = 2y,dy$. With this substitution $u$ changes from 0 to 1 and we integrate at the start {align *} int_0^1int_0^ye^{y^2} dx,dy &=int_0^1 ye^{ y ^ 2} dy & = int_0 ^ 1 frac {1} {2} e ^ {u} du = frac {1} {2} e ^ {u} Big | _0 ^ 1 = frac {1} {2} (e-1 ). end {align *} Example 3Read more: how to mix yellow paint with base color This is a slightly more complicated example. Reverse the order of integration in the following integral. begin {collect *} int_ {pi / 2} ^ {5pi / 2} int_ {sin x} ^ 1 f (x, y) dy, dx end {collect *} Solution: The region $dlr$ is described by the product This fraction is the start {set *} pi / 2 le x le 5pi / 2 sin x le y le 1. the end {set *} as shown in the following image, where the total range is over $ x$ is shown by the gray bar below the region, and the change boundaries for y$ are shown by the blue and cyan curves.
One trick for varying variables with this region is to correctly handle the lower boundary $y = sin(x)$. When we solve this boundary equation for x as a function of y, we might be tempted to write it x = arcsin(y) and even think x le arcsin(y) $ in the region .Looking closely at the image, we see that this cannot happen. In fact, the lower bound of y as a function of x (the blue curve) should be both the upper bound and the lower bound of x as a function of y, as given Shown in red and purple the curves in the image below.
To get the formula for these boundaries, we must remember how the inverse of the sine, $arcsin(y)$, is determined. To determine the inverse of sin(x)$, we need to limit the function to the amount of time it takes on each value only once. The standard way to define $arcsin(y)$ is to constrain $sin(x)$ to $x values in the period $[-pi/2,pi/2]$as $sin(x)$ ranges from $-1$ to 1 in that period. This means that $arcsin(y)$ is between $[-pi/2,pi/2]$ as $ y $ goes from $-1 $ to 1. For the upper boundary of $ x $ (purple), $ x $ is between $3pi / 2 $ to $5pi / 2 $. If we put x = arcsin(y) + 2pi$, then x = 3pi/2 when y = -1$ and x = 5pi/2 when y = 1$, as required. For the lower boundary of $x$ (red), we need $x$ to be a decreasing function of $y$, starting at $x = 3pi/2 when $y = -1$ and decreasing to $ x = pi / 2 $ when $ y = 1 $. These conditions are satisfied if we choose $x = pi-arcsin(y)$. If you are an expert on your trigonometric identity, you can verify that the equations for both of these curves are just different inverses of sin(x), because taking the sine of the equations This program reduces them to $y = sin(x)$. Since within the area, $y $ fluctuates over the period $[-1,1]$ (gray bar to the left of the region), we can describe the region $dlr $ with inequalities starting {set *} -1 le y le 1 pi-arcsin y le x le arcsin y + 2pi. end {collect *} This $dlr$ description is what we need to change the integration order and we see that begin {collect *} 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. end {collect *} More Examples If you want more examples of double integrals, you can study some introductory double integral examples.. You can also see examples of integrals. double from special cases when interpreting double integrals as areas and double integrals as volume.Read more: Mordhau: how to prevent success in combat
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