What equations cant you integrate? (1 Viewer)

Buiboi

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What equations cant you integrate? but this does not involve like trig or logs just integration in general...like you know how some graphs you ahve to find the area below it between curve and x-aixs and y axi-s, sometimes you cant integrate the equation so you find the area another way....what is the other way and how can you tell when you cant integrate, like besides simpsons rule and trapezoidal
 
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pLuvia

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Well, there are some graphs that are plotted below and above the x axis, and when trying to find the area you have to actually see where the graph is below the x axis and above the x axis to make sure you don't miss out on the graph below the x axis

From what I've encountered only the log graphs need the other method of integrating (within the 2u/3u level) while at 4u level you learn a method to integrate lnx, lnxsinx and the like
 

Buiboi

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are you talking about using absolute values to find area because its positioned in a negative spot so the absolute value basically converts it back to positive?
 
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pLuvia

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Buiboi said:
are you talking about using absolute values to find area because its positioned in a negative spot so the absolute value basically converts it back to positive?
Precisely
 

toadstooltown

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Well, some areas you might be asked to find are, say, under arcsine or arccos graphed [sin^(-1)x etc ] or log graphs. To find the area here them you find the area to the y-axis and subtract it from the whole rectangle, if you kinda get me. You can integrate sinx and cosx so the area will be the whole area for the upper limit to both axes - the area you can integrate. Hard to explain, but you'll get it.

Another type of integration often occurs when they give you something particularly nasty or muddled to differentiate (usually a 'show' question) and then the next part of the question has the function as a slight variation of the derivative so you know what the integral would be (what you initially differentiated but amended)

Basic Example.
i)(d/dx) x.lnx-x
=x×1/x+1×lnx - 1
=1-1+lnx
=lnx
ii) hence find the area covered by a particle from a to be seconds if it's velocity is given by ln2x where x is seconds.

You couldn't previously integrate ln2x but now you know what it came from so you can. These only occur every so often but are good to know. If you do maths in the future then it's good to note weird differentiations especially for hard things like arctan quickly.
 
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