Partial integrals or partial integration could mean several different things. On this site, we use the term to refer to integrals with multiple variables usually involving nested integrals. These types of integrals are often called iterated integrals. Other meanings could be integration by parts or integration using partial fractions.
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A partial integral is the complement of the partial derivative. The idea is that your integral has multiple variables but you integrate with respect to one at a time, while holding the other variables constant. The idea is not difficult if you have a good handle on partial derivatives. There is one additional thing you need to watch for, the form of the constant.
You remember way back in single variable calculus when you learned integration, you needed to remember to include the constant of integration. For example,
\( \int{2x~dx} = x^2 + C \)
In this example, the \(C\) is just a constant, i.e. just some real number. We use \(C\) as a placeholder, since it can have any value, \(3\), \(\pi\), \(17\), \(127\) and even \(0\).
The same idea holds for partial integrals, i.e. you need a 'constant' of integration. However, this constant is a bit different for multi-variable integration than it was for the single variable case. The 'constant' of integration here is actually a constant function of integration, i.e. it is a function of the other variables that are held constant when we integrate.
In general, it looks like this. For a function \(f(x,y)\), when we integrate this with respect to \(x\), we get
\( \int{f(x,y)~dx} = F(x,y) + g(y) \) where \(\partial F(x,y)/\partial x = f(x,y)\). Let's think about this.
If we were given a function that looks like \(F(x,y) + g(y)\) and we were told to take the partial derivative with respect to \(x\), we would get
\(\displaystyle{ \frac{\partial [F(x,y)]}{\partial x} + \frac{\partial [g(y)]}{\partial x} = f(x,y) + 0 }\)
\(\displaystyle{ \frac{\partial [g(y)]}{\partial x} = 0 }\) because \(g(y)\) has no \(x\)'s in it and, therefore, is considered a constant. And we all know that the derivative of a constant is zero.
So, other than the constant situation, integration should be pretty straightforward. Let's do an example.
Evaluate \( \int{ 2xy~dx } \)
Evaluate \( \int{ 2xy~dx } \)
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Since we are integrating with respect to \(x\), \(y\) is a constant. In fact, when integrating, we can pull the \(y\) outside the integral to make things easier to see. So we have
\( \int{ 2xy~dx } = y \int{2x~dx} = y(x^2) + g(y) = x^2y + g(y) \)
Notice that instead of \(C\), we have \(g(y)\), which is a function of \(y\). (This function could also be just a constant.) The point is, it can't have any \(x\)'s in it. It can have only constants and \(y\)'s (or whatever variable(s) we hold constant during integration).
Here is a great video clip with an example. He explains the details without getting too technical.
video by PatrickJMT |
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Definite Partial Integrals
For definite integrals, we don't have worry about the constant, of course. However, we have an added tip that may help you avoid mistakes when you are tired or stressed, for example, during an exam. Let's look at an example.
Evaluate \(\displaystyle{ \int_{a}^{b}{ 2xy ~ dx } }\)
Evaluate \(\displaystyle{ \int_{a}^{b}{ 2xy ~ dx } }\)
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\(\displaystyle{ \int_{a}^{b}{ 2xy ~ dx } = }\) \(\displaystyle{ 2y \int_{a}^{b}{ x~dx } = }\)
\(\displaystyle{ \left[ x^2 y \right]_{x=a}^{x=b} = }\) \(\displaystyle{ (b^2-a^2)y }\)
Most of this example is self-explanatory but we want to mention one thing we did that may look unusual. Notice that after integration but before substituting the limits of integration, we wrote \(\displaystyle{ \left[ x^2 y \right]_{x=a}^{x=b} }\). It would have been technically correct if we had written \(\displaystyle{ \left[ x^2 y \right]_{a}^{b} }\). However, writing \(x=a\) and \(x=b\), we are making sure that we remember to substitute for \(x\), not \(y\). It is very easy to make a mistake here. So writing it more clearly could help prevent lost points. We highly recommend that you do it this way but, as always, check with your instructor to see what they expect.
Another twist with definite partial integrals is that the limits of integration do not have to be constants. They can be expressions like \(y+2\). This doesn't really change anything that you will do but it looks kind of strange when you first see it. Let's look at another example.
Evaluate \(\displaystyle{ \int_{0}^{x}{2x+y ~ dy} }\)
Evaluate \(\displaystyle{ \int_{0}^{x}{2x+y ~ dy} }\)
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\(\displaystyle{ \int_{0}^{x}{2x+y ~ dy} = }\) \(\displaystyle{ \left[ 2xy + y^2/2 \right]_{y=0}^{y=x} = }\)
\(\displaystyle{ 2x^2 + x^2/2 = 5x^2/2 }\)
A few comments are in order.
1. Notice that we used the explicit notation for \(y=0\) and \(y=x\) for the limits. We do that a lot on this site to help you follow our examples.
2. If you have an expression in one or both limits of integration, the expressions will not contain the variable of integration under any circumstance. Here is something you will never see: \(\displaystyle{ \int_{0}^{y}{2x+y ~ dy} }\). Having a limit of integration with the variable of integration in it does not make sense.
3. Notice that the result of the integration is a function of the variable that was considered a constant in the integration and the variable of integration is no where to be found. This will always be the case. In this example, \(y\) is not in the final answer and \(x\) was considered a constant of integration but in the final answer it is a variable.
We usually don't use partial integration in isolation. The beauty of integration is that we use this idea to solve double and triple integrals. Here are a few practice problems. After those, if you are comfortable describing areas in the plane, we recommend that you go straight to your next topic, double integrals on rectangular regions.
Practice
Unless otherwise instructed, evaluate these integrals using correct notation. Give your answers in exact terms, completely factored. Do not forget the function of integration.
\(\displaystyle{ \int_{1}^{2x}{ \frac{x^2}{y} ~ dy } }\)
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\(\displaystyle{ \int_{1}^{2x}{ \frac{x^2}{y} ~ dy } }\)
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\( 2x^2 \)
Problem Statement |
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\(\displaystyle{ \int_{1}^{2x}{ \frac{x^2}{y} ~ dy } }\)
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Remember that since we are integrating with respect to \(y\), \(x\) is considered a constant when we integrate.
\(\begin{array}{rcl} \int_0^x{ x + 2y ~ dy } & = & \left[ xy + y^2 \right]_{y=0}^{y=x} \\ & = & [x^2 + x^2] - [0] \\ & = & 2x^2 \end{array}\)
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\( 2x^2 \)
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\(\displaystyle{ \int_{x}^{x^2}{ \frac{y}{x} ~ dy } }\)
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\(\displaystyle{ \int_{x}^{x^2}{ \frac{y}{x} ~ dy } }\)
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\(\displaystyle{ \frac{x}{2}[ x^2 - 1 ] }\)
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\(\displaystyle{ \int_{x}^{x^2}{ \frac{y}{x} ~ dy } }\)
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In this integral, \(y\) is the variable of integration, so \(x\) is a constant. Before integrating, we will pull the \(1/x\) out in front (since it is a constant). This makes the integration a bit easier to see.
\(\begin{array}{rcl} \displaystyle{ \int_{x}^{x^2}{ \frac{y}{x} ~ dy } } & = & \displaystyle{ \frac{1}{x} \int_{x}^{x^2}{ y ~ dy } } \\ & = & \displaystyle{ \frac{1}{x} \left[ \frac{y^2}{2} \right]_{y=x}^{y=x^2} } \\ & = & \displaystyle{ \frac{x^4}{2x} - \frac{x^2}{2x} } \\ & = & \displaystyle{ \frac{x^3}{2} - \frac{x}{2} } \\ & = & \displaystyle{ \frac{x}{2} [ x^2 -1 ] } \end{array}\)
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\(\displaystyle{ \frac{x}{2}[ x^2 - 1 ] }\)
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\(\displaystyle{ \int_{0}^{\cos y}{ y ~ dx } }\)
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\(\displaystyle{ \int_{0}^{\cos y}{ y ~ dx } }\)
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\( y \cos y \)
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\(\displaystyle{ \int_{0}^{\cos y}{ y ~ dx } }\)
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\( \displaystyle{ \int_{0}^{\cos y}{ y ~dx } = yx |_0^{\cos y} = y \cos y } \)
Final Answer |
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\( y \cos y \)
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\(\displaystyle{ \int_{e^y}^{y}{ \frac{y \ln x}{x} ~ dx } }\), \( y \gt 0 \)
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Evaluate \(\displaystyle{ \int_{e^y}^{y}{ \frac{y \ln x}{x} ~ dx } }\), \( y \gt 0 \)
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\(\displaystyle{ \frac{y}{2}[ (\ln y)^2 - y^2 ] }\)
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Evaluate \(\displaystyle{ \int_{e^y}^{y}{ \frac{y \ln x}{x} ~ dx } }\), \( y \gt 0 \)
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Use integration by substitution and let \( u = \ln x \to du = (1/x)dx \to x~du = dx \) |
Drop the limits of integration and evaluate the indefinite integral \(\displaystyle{ \int{ \frac{y \ln x}{x} ~ dx } }\) |
\(\displaystyle{ \int{ \frac{y}{x}u ~ x~ du } = y \int{ u ~ du} = \frac{y u^2}{2} }\) |
Now convert the integral back in terms of \(x\) and apply the limits of integration. |
\(\displaystyle{ \left[ \frac{y}{2}(\ln x)^2 \right]_{x=e^y}^{x=y} }\) |
\(\displaystyle{ \frac{y}{2}\left[ (\ln y)^2 - (\ln e^y)^2 \right] }\) |
\(\displaystyle{ \frac{y}{2} [ (\ln y)^2 - y^2 ] }\) |
If you converted the limits of integration in terms of \(u\), here are the steps.
Use integration by substitution and let \( u = \ln x \to du = (1/x)dx \to x~du = dx \) |
For the limits of integration, we have the upper limit \(x=y \to u =\ln y \). For the lower limits \(x=e^y \to u = \ln e^y \to u = y \) |
\(\displaystyle{ \int_{u=y}^{u=\ln y}{ \frac{y}{x}u ~ x~ du } = y \int_{u=y}^{u=\ln y}{ u ~ du} = \left[ \frac{y u^2}{2} \right]_{u=y}^{u=\ln y} }\) |
Now apply the limits of integration. |
\(\displaystyle{ \frac{y}{2} [ (\ln y)^2 - y^2 ] }\) |
Final Answer |
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\(\displaystyle{ \frac{y}{2}[ (\ln y)^2 - y^2 ] }\)
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For \( f(x,y) = 3x^3 - 4/y^2 -x + xy + 2y\), calculate the partial derivatives \(f_x\) and \(f_y\), then integrate each using the same variable, i.e. evaluate \( \int{ f_x ~ dx } \) and \( \int{ f_y ~ dy } \)
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For \( f(x,y) = 3x^3 - 4/y^2 -x + xy + 2y\), calculate the partial derivatives \(f_x\) and \(f_y\), then integrate each using the same variable, i.e. evaluate \( \int{ f_x ~ dx } \) and \( \int{ f_y ~ dy } \)
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video by MIP4U |
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\(\displaystyle{ \int_{0}^{2y}{ 2x - 3y ~ dx } }\)
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\(\displaystyle{ \int_{0}^{2y}{ 2x - 3y ~ dx } }\)
Solution |
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video by MIP4U |
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\(\displaystyle{ \int_{1}^{2x}{ \frac{x^2}{y} ~ dy } }\)
Problem Statement |
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\(\displaystyle{ \int_{1}^{2x}{ \frac{x^2}{y} ~ dy } }\)
Solution |
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video by MIP4U |
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Really UNDERSTAND Calculus
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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Practice Instructions
Unless otherwise instructed, evaluate these integrals using correct notation. Give your answers in exact terms, completely factored. Do not forget the function of integration.