Not understanding some integration results

  • #1

Main Question or Discussion Point

I was wondering if you guys could help me understand something. I have an equation that, when I graph it, looks a lot like a regular sine function. However, when I integrate it to find the area under it and graph the new function, I'm getting a negative result in the first 0.56 radians or so.

The equation is:

dy = xSIN(x) - 0.03356(x^2)SIN(x) dx

Integrating for x (actually, letting Wolfram Mathematica Online Integrator do it for me), I get:

y = (1 - 0.06712x)SIN(x) + 0.03356(x - 29.8643)(x + 0.0669695)COS(x)

(I'm trying to analyze this for x = 0 to pi, by the way.)

The first equation gives me nothing negative, so why when it's integrated and graphed do I get a negative dip before x = 0.56?
 

Answers and Replies

  • #2
gb7nash
Homework Helper
805
1
Unless you have a given initial condition for y (i.e. y(a) = b), you cannot graph the indefinite integral just knowing the area under the graph. Wolfram alpha gives you a + Constant at the end, not sure why you got rid of that. However, if the problem you're working on does have an initial condition, then you can use that to find the constant and graph the integral properly.
 
  • #3
Wolfram didn't mention a constant, though I know I could use one. I just didn't think I needed to. ... I'm including two pictures here - screenshots of an online graphing calculator (http://www.coolmath.com/graphit/) where I plugged in these equations. In both of them, I've zoomed into the region of x = 0 to about 3.14.View attachment Capture1.bmp View attachment Capture2.bmp The first picture is a graph of the original equation. As you can see, the curve is entirely in the y+ quadrant. But in the second picture, after the equation is integrated, the first part of the curve is negative. It's been a long time since I took a calculus class, but isn't an integration supposed to give me the area under the curve? And when I solve the integrated equation for, let's say, x = 0.5, shouldn't that give me the area under the curve down to the y = 0 line, and from x = 0 to x = 0.5? But when I solve for x = 0.5, I get -0.027. That makes sense given the second picture, but it doesn't make sense to me given the first picture. Where's the negative area under the curve in the first graph?
 
  • #4
gb7nash
Homework Helper
805
1
I think you're getting definite integrals confused with indefinite integrals. If you're dealing with a definite integral (i.e. an integral with a lower bound and upper bound), you will get the exact area under the curve between the two x values. However, if you find the indefinite integral, you cannot graph it unless you know what the constant is. The program you used might have just set a default value for a constant, maybe 0.

Assuming you know what the constant is and looking at the two graphs you have, it's usually easier looking at the slope instead of area. The slope of the second graph should be the y-value of the first graph.
 
  • #5
Bingo. If I add a constant, bringing y up to 0 when x = 0, that solves my problem, and the rest of the numbers line up with what I thought they should be. Thanks a lot!
 

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