A quiz for Max and Seva

by Rishi Mon Apr 22, 2013 8:11 am

Write an integral representing the area of the region bounded by the lines x=3, x=4, the function f(x)=((8⋅x)+5),
and the function g(x)=arctan(cos(x)).

Man. I am having a lot of fun doing these kinds of problems.

by Seva Lamberdar Mon Apr 22, 2013 8:26 am

Rishi wrote:
Man. I am having a lot of fun doing these kinds of problems.

Good for you.

by MaxEntropy_Man Mon Apr 22, 2013 9:51 am

integrate dx. dy with the following limits:

y= arctan(cos(x)) to 8.x+5, and

x= 3 to 4

good you are having fun.

by Rishi Mon Apr 22, 2013 11:25 am

MaxEntropy_Man wrote:integrate dx. dy with the following limits:

y= arctan(cos(x)) to 8.x+5, and

x= 3 to 4

good you are having fun.

Is it the same thing as

Integral of abs((8x+5) -(arctan(cos(x))) dx limits 3 to 4 ?

by Hellsangel Mon Apr 22, 2013 11:30 am

Are you getting the poor Suchkers to do your assignments?

by Rishi Mon Apr 22, 2013 11:51 am

MaxEntropy_Man wrote:integrate dx. dy with the following limits:

y= arctan(cos(x)) to 8.x+5, and

x= 3 to 4

good you are having fun.

I am not yet into double integrals.

My answer is correct.

I wonder if the double integral stuff you posted takes care of the absloute value.

by Seva Lamberdar Mon Apr 22, 2013 11:53 am

Remember you want the area to be always positive, hence the issue of absolute value.

by Jeremiah Mburuburu Mon Apr 22, 2013 2:44 pm

Rishi wrote:
MaxEntropy_Man wrote:integrate dx. dy with the following limits:

y= arctan(cos(x)) to 8.x+5, and

x= 3 to 4

good you are having fun.

I am not yet into double integrals.

My answer is correct.

I wonder if the double integral stuff you posted takes care of the absloute value.

i don't think it does, but in this case, max is right because, for every x in the interval 3 <= x <= 4, 8x + 5 lies above arctan(cos(x)), i.e., 8x + 5 - arctan(cos(x)) is positive for every x in that interval, and in particular, does not change sign. note that max's upper limit in the integration over y-values is 8x + 5.

your integral is correct, and because of the absolute value, is less risky than max's. if you were to find and evaluate the integral symbolically, i think you'd need to know at what value(s) of x, if any, the integrand changes sign.

the numerical value of the integral is about 33.7286.

caution: i haven't done this sort of thing in a long time.

by MaxEntropy_Man Mon Apr 22, 2013 3:56 pm

rishi -- your answer and mine are identical. i am just used to writing areas as double integrals.

also JM is right, but it does not hurt to enclose the integral in an absolute value sign just to be sure.

i did a numerical integration using mathematica and my numerical answer is identical to JM's.

by MaxEntropy_Man Mon Apr 22, 2013 4:13 pm

you don't even need a copy of mathematica. you can just do this on wolfram alpha gratis:

http://www.wolframalpha.com/input/?i=Integrate[Integrate[1%2C+{y%2C+ArcTan[Cos[x]]%2C+8+x+%2B+5}]%2C+{x%2C+3%2C+4}]

by Seva Lamberdar Mon Apr 22, 2013 4:14 pm

The way problem is defined, the double integral has nothing to with absolute value. The absolute value is there to get the final answer (area in this case) a positive number, whether going from x = 3 to x =4, or from x = 4 to x= 3, or from y = arctan(cos(x)) to y = 8.x+5, or from y = 8.x+5 to y = arctan(cos(x)). Negative area makes no sense anyway unless someone is in vector calculus and the area normal (whether pointing inward or outward) matters.

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