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Application of Integration

Application of Integration

by Prashanth S -
Number of replies: 3
Find the area bounded by the curve y = 2x + x2 - x3 , the x-axis and the lines x=-1 and x=1
The graph of this curve is like a sinusoidal curve . In the –ve x-axis the curve lies below the x-axis and in the +ve x-axis it lies above . my doubt is if the curve lies below the x-axis should we take negative of the function ? How do we solve this problem ?
In reply to Prashanth S

Re: Application of Integration

by Bharath krishna -

This graph is not sinusoidal as it is not periodic.To solve this problem first integrate the function from -1 to 0.This will give a -ve ans as the curve lies below the x-axis between -1 and 0.Take modulus of this value obtained.Now integrate the function from 0 to 1 and add the resulting +ve answer to the "+ve" answer obtained previously by taking mod.This will give the reqrd answer. If you directly integrate from -1 to 1, you will get the difference in the areas bounded by the +ve part of the graph and -ve part of it and not the actual area.

In reply to Prashanth S

Re: Application of Integration

by Prashanth S -
Iam not saying the graph is Sinusoidal . iam just trying to say that the graph looks like a sinusoid . the function is not periodic.
my doubt is whether we take it as
($ ----> integration )


0 1
$ -(2x+x²+x^3)+ $ (2x+x²+x^3)
-1 0
In reply to Prashanth S

Re: Application of Integration

by Bharath krishna -
Yes, that is right. if we integrate the -ve part of the graph, by adding a -ve sign to function, u will get a +ve answer which is correct.It is equivalent to taking modulus after integration.