FR0568 #7

Forum for the GRE subject test in mathematics.
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Joined: Fri Mar 26, 2010 7:18 am

FR0568 #7

Postby thmsrhn » Sat Mar 27, 2010 4:58 am

Hey does one integrate the modulus of a function?

Here s the question:

integrate the mod of (X+1)dx whose upper and lower limits are 3 and -3 respectively.

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Joined: Fri Oct 23, 2009 11:42 pm

Re: FR0568 #7

Postby origin415 » Sat Mar 27, 2010 11:27 am

the modulus/absolute value is just
|a| = \begin{cases} a &a \geq 0 \\ -a &a < 0 \end{cases}

So your function |x+1| can actually be split into two normal functions, without any absolute values. Add the results of integrating them separately.

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Joined: Sun Sep 05, 2010 11:11 am

Re: FR0568 #7

Postby mdornbos » Sun Sep 05, 2010 11:17 am

The way I did this problem was to simply graph the function |x+1| from -3 to 3. You will notice that integrating this function creates two triangles. Then use the formula for area of a triangle and add up the two separate triangles areas to get the answer. I believe its (1/2*2*2) which is the triangle from x=-3 to x=-1, and (1/2*4*4) which is the triangle from x=-1 to x=3. Adding up the two areas you get 2+8=10.

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