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Definite Integral = Area Under curve(s) 
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#1
Feb1609, 06:44 AM

P: 125

I know the title can be a bit misleading, yet it's really close to what I want to ask.
Today I came upon an integral at school. The really easy one :[itex]\int_{1}^{1}x^3 dx[/itex]. Of course, calculating the integral, you get 0. Yet, as far as I know, the way to get the area under the curve is by integrating the function inside the limits of which the area above or under x you want to get. Yet [itex]f(x)=x^3[/itex] occupies some space from 1 to 0, and the same space from 0 to 1. Why then does the integral consider the one being positive space and the other one being negative, sum them up and calculate the area as zero? This is a question that I've had before, and by looking I saw that there are restrictions or,lack of a better word, modifications that must be applied to the integral when calculating for the area under the curve. Could you please introduce me to them? Thanks in advance! 


#2
Feb1609, 07:13 AM

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PF Gold
P: 16,091

Well, you should be able to get an answer yourself, just by going back to the definition of the definite integral, and reviewing the argument that indicates how it relates to the area under a curve.
There is a deeper idea that areas in a plane, or volumes in 3space can be oriented, just like segments in a line... and reflections reverse orientation. When you're thinking of the area under a curve, what you're really thinking is the area between the xaxis and your curve... and you can orient that direction: you can think of it being from the xaxis to the curve. When paired with an orientation on the xaxis (we integrate from left to right), that gives us a twodimensional orientation on the area we're integrating. And you'll notice that when the curve negative, the direction 'from' the xaxis 'to' the curve is opposite than when the curve is positive.... 


#3
Feb1609, 07:14 AM

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Thanks
PF Gold
P: 39,565

A little more "sophisticated" problem would be to find the area between the graphs of y= f(x) and y= g(x) when the graphs cross one or more times. In that case you would have to integrate f(x) g(x) so that, no matter which graph was the upper bound, the integral would be positive. 


#4
Feb1609, 07:51 AM

P: 125

Definite Integral = Area Under curve(s)
Nice quoting the "Thanks in advance" there :)
To the point now. First of all, I understand that there is no such thing as positive or negative space and length, but that was my initial thought on what the integral "thought it should be doing", if you understand what I am trying to say. I see in my textbook that for functions that also have negative values, we use [itex]\int_{a}^{b}f(x)dx[/itex], instead of [itex]\int_{b}^{a}f(x)dx [/itex]. But doesnt that mean we are still going to have [itex]\int_{1}^{1}f(x)dx[/itex] , or are we going to take [itex]\int_{1}^{1}x^3dx = [x^4/4]^{1}_{1}=1/2[/itex], or [itex]\int_{1}^{1}x^3dx = [x^4/4]^{1}_{1}=?[/itex](1/2 again??) Basically, I am sort of confused. You mentioned something about a lower limit. Does this mean I should be calculating the integral (asking for area between curve and x'x axxis) by cutting the initial integral into two separate integrals? Like this? : [itex]\int_{1}^{1}x^3dx=\int_{1}^{0}x^3dx + \int_{0}^{1}x^3dx?[/itex] 


#5
Feb1609, 08:17 AM

P: 810

If you want the actual area instead of a signed area, you must take the integral of the absolute value of the function you want. The absolute value flips all the negative parts of the graph so that the negative areas magically turn into positive ones. 


#6
Feb1609, 08:58 PM

P: 1

1. (i) [tex]v=u+at[/tex]



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