hotcommodity Posted July 25, 2007 Share Posted July 25, 2007 I have a few questions, one concerning a basic integral that I'm having trouble working out, one about notation, and lastly about graphing a two dimensional function in R^3. This is the integral I'm having trouble with: [math] \int e^{2x} sin(3x) dx [/math] I'm pretty sure I have to start off by using u-substitution, and then use integration by parts. I think the u-substitution is what's giving me trouble. If I let [math] z= 3x [/math] then [math] dz= 3 dx [/math] and therefore [math] \frac{1}{3}dz = dx [/math]. If [math] z= 3x [/math] then [math] x= \frac{z}{3}[/math]. Then I would have [math] \frac{1}{3}\int e^{\frac{2z}{3}} sin(z) dz [/math] Is this correct? Secondly, I was wondering if someone could tell me what the "d" in dx means. I know that the derivative is [math] \frac{\Delta y }{\Delta x} [/math] as [math]\Delta x[/math] goes to zero. So does [math]\frac{dy}{dx}[/math] always imply that the change in x goes to zero? Why is "dx" in integration expressions? Lastly, i was playing around with my TI-89, and decided to graph y=sin(x) in 3D. It came up as a plane instead of a line. I guess I'm just having a hard time picturing why that is. Any help is appreciated. Link to comment Share on other sites More sharing options...

clarisse Posted July 26, 2007 Share Posted July 26, 2007 I have a few questions, one concerning a basic integral that I'm having trouble working out, one about notation, and lastly about graphing a two dimensional function in R^3. This is the integral I'm having trouble with: [math] \int e^{2x} sin(3x) dx [/math] I'm pretty sure I have to start off by using u-substitution, and then use integration by parts. I think the u-substitution is what's giving me trouble. If I let [math] z= 3x [/math] then [math] dz= 3 dx [/math] and therefore [math] \frac{1}{3}dz = dx [/math]. If [math] z= 3x [/math] then [math] x= \frac{z}{3}[/math]. Then I would have [math] \frac{1}{3}\int e^{\frac{2z}{3}} sin(z) dz [/math] Is this correct? Secondly, I was wondering if someone could tell me what the "d" in dx means. I know that the derivative is [math] \frac{\Delta y }{\Delta x} [/math] as [math]\Delta x[/math] goes to zero. So does [math]\frac{dy}{dx}[/math] always imply that the change in x goes to zero? Why is "dx" in integration expressions? Lastly, i was playing around with my TI-89, and decided to graph y=sin(x) in 3D. It came up as a plane instead of a line. I guess I'm just having a hard time picturing why that is. Any help is appreciated. While your U-substitution is fine (and so is your resulting integral) I don't see it getting you very far. If I were you I would do integration by parts. If you have any problems you can ask but it is fairly simple if you remember to be consistent in your choice of u and dv/dx. The dx, simply refers to "change in" x I think, as you know that a derivative is a rate of change so dy/dx simply means the change in y over the change in x. And I think that the reason the sine function graphed in 3d appears as a plane, is because x is a plane itself because y= sinx would be true for all points in the z-plane. (I don't know if I worded that clearly enough). Link to comment Share on other sites More sharing options...

ajb Posted July 28, 2007 Share Posted July 28, 2007 I agree with your substitution in the integral. Now you are going to use the "parts formula"? Without going into measure theory (which I don't know that much about), I like to think of "dx" as meaning an infinitesimal change in x. Then this makes sense in Riemann's definition of integration. Link to comment Share on other sites More sharing options...

river_rat Posted July 29, 2007 Share Posted July 29, 2007 or if you feel daring, your answer is [math] \Im \left( \int e^{(2+3i) x} dx \right) [/math] which saves you all the product rule pain PS [math] \Im [/math] denotes the imaginary part if you are wondering. The [math] dx [/math] is called a differential form (or one form in this case) and the theory here is quite interesting. It takes a surprising amount of mathematical work to get something that is more meaningful then the nonsense idea of an infinitely small but non-zero change in x. Link to comment Share on other sites More sharing options...

hotcommodity Posted July 30, 2007 Author Share Posted July 30, 2007 I appreciate the replies above. Well I worked out the integral using integration by parts only, and I got an answer much different from the books. I let u= sin(3x), du= 3cos(3x), dv= e^(2x), and v= e^(2x). This gives me: [math] \int e^{2x} sin(3x) dx= sin(3x)*e^{2x} - 3 \int cos(3x)*e^{2x} dx[/math] Then I use integration by parts a second time, with u= cos(3x), du= -3sin(3x), dv= e^(2x), and v= e^(2x). This gives me: [math] \int e^{2x} sin(3x) dx=sin(3x)*e^{2x} - 3cos(3x)*e^{2x}- 9 \int sin(3x)*e^{2x} dx [/math] I figured that since I have the same integral on both sides, I can just add 9 of that integral to each side, then divide the right side by 10, which gives: [math] \int e^{2x} sin(3x) dx= (1/10)*(sin(3x)*e^{2x} - 3cos(3x)*e^{2x} + C)[/math] So to check my answer I made the integral a definite integral, and checked each answer using a program on my calculator. Neither the books answer or my answer appears to be correct. Also, about the "dx" in integration expression, I'm still not grasping the concept that easily, but I'll come back when I have more time and question what I'm having trouble with in more depth. Link to comment Share on other sites More sharing options...

clarisse Posted July 31, 2007 Share Posted July 31, 2007 I appreciate the replies above. Well I worked out the integral using integration by parts only, and I got an answer much different from the books. I let u= sin(3x), du= 3cos(3x), dv= e^(2x), and v= e^(2x). This gives me: [math] \int e^{2x} sin(3x) dx= sin(3x)*e^{2x} - 3 \int cos(3x)*e^{2x} dx[/math] Then I use integration by parts a second time, with u= cos(3x), du= -3sin(3x), dv= e^(2x), and v= e^(2x). This gives me: [math] \int e^{2x} sin(3x) dx=sin(3x)*e^{2x} - 3cos(3x)*e^{2x}- 9 \int sin(3x)*e^{2x} dx [/math] I figured that since I have the same integral on both sides, I can just add 9 of that integral to each side, then divide the right side by 10, which gives: [math] \int e^{2x} sin(3x) dx= (1/10)*(sin(3x)*e^{2x} - 3cos(3x)*e^{2x} + C[math] So to check my answer I made the integral a definite integral, and checked each answer using a program on my calculator. Neither the books answer or my answer appears to be correct. Also, about the "dx" in integration expression, I'm still not grasping the concept that easily, but I'll come back when I have more time and question what I'm having trouble with in more depth. I didn't read it through but I think that if you're not getting the correct answer it's because you're getting the integral of [math]e^(2x)[/math] incorrectly. Remember the chain rule, so if you differentiate what you got for by integrating [math]\frac{dv}{dx} [/math], [math]v = e^(2x))[/math] you would get [math]2e^(2x)[/math] so to cancel out that extra two just divide it by two so your [math]v[/math] should equal [math]\frac{e^2x}{2} [/math] Link to comment Share on other sites More sharing options...

hotcommodity Posted July 31, 2007 Author Share Posted July 31, 2007 Thanks for the reply, you were right. I was integrating e^(2x) wrong. At least this way I got the same answer that the book had. Link to comment Share on other sites More sharing options...

river_rat Posted August 6, 2007 Share Posted August 6, 2007 I'm surprised no one chastised me for my solution - oh well Link to comment Share on other sites More sharing options...

BenTheMan Posted August 11, 2007 Share Posted August 11, 2007 don't you need to turn it into a contour integral or something? Link to comment Share on other sites More sharing options...

Country Boy Posted August 16, 2007 Share Posted August 16, 2007 As far as your last question, "Lastly, i was playing around with my TI-89, and decided to graph y=sin(x) in 3D. It came up as a plane instead of a line. I guess I'm just having a hard time picturing why that is." is concerned, why would you expect it to be a line? If you were to graph x= 0 in 1D (i.e. a number line) it would be a single point. In 2D, it would be a vertical line (the y-axis), because each point has both x and y coordinates now and "x= 0" allows y to be anything. In 3D, "x= 0" is a plane (the yz-coordinate plane) because both y and z can be anything. Another way of looking at it is this: Selecting an arbitrary point in 3 dimensions you have 3 "degrees of freedom"- you can choose any 3 numbers you want. If you have one equation, f(x,y,z)= 0, say, then you can (theoretically) choose any two numbers for two variables and solve for the third variable. You have two "degrees of freedom" and the graph is a surface. (And, of course, the graph of y= sin(x) is a general surface, NOT a "plane".) Link to comment Share on other sites More sharing options...

K!! Posted January 2, 2008 Share Posted January 2, 2008 or if you feel daring, your answer is [math] \Im \left( \int e^{(2+3i) x} dx \right) [/math] which saves you all the product rule pain Actually, this is the nice method to tackle such integrals. It takes less time. Link to comment Share on other sites More sharing options...

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