Integral displays infinity (mistake ?)

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Posted on 2008-08-17 06:10 By KewlEugene (

20)

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f := sec(x)^2;
Int(f, x = 0 .. 5*Pi*(1/4)) = int(f, x = 0 .. 5*Pi*(1/4))

I enter the above in Mape 12 Student Edition, and the answer is infinity, but it's obviously 1.

How can I get Maple to display the obvious ?

singularity

Comment on 2008-08-17 06:55 from Axel Vogt ( Maple Rating 4

948)

if you would use common notations, then your f = 1/(cos(x)^2),
so at Pi/2 ... hence you should not write down the integral

Comment on 2008-08-17 07:44 from KewlEugene (

20)

what are 'comon notations' ?

can you show me the code that results ina carrorect answer whic is 2.

Comment on 2008-08-17 09:35 from Axel Vogt ( Maple Rating 4

948)

plot(1/(cos(x)^2), x=0 .. 5*Pi*(1/4), y= 0..6);

Comment on 2008-08-17 11:06 from Alex Smith ( Maple Rating 3

289)

You are probably thinking the answer is 1 because the antiderivative is tan(x), and

tan(5*Pi/4)-tan(0)=1.

But the integrand f(x)=sec(x)^2 is neither bounded nor continuous over the interval, so the fundamental theorem of calculus does not apply.

Comment on 2008-08-19 08:34 from KewlEugene (

20)

ok. i tried the upper bound of pi/4 and that resulted in 2. so... thanks guys...

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