Are Definite Integrals and Area Calculations Performed Similarly?

[Jacobpm64]

ok, my question involves two different sets of directions ..

1. Use integration to find the area of each shaded region.
2. Evaluate each definite integral.

Ok, my question is this... do i do the same thing for both of these directions? ..

Even further... say I have a function that goes over and under the x-axis.. and I'm asked to find the area when the function is bounded in such a way that the area would be both over and under the x-axis... For direction #1, i would find the zeros of the function and use the fundamental theorem of calculus and do each part separately.. taking the absolute value of each.. and adding the two areas together.. ok.. i got that.. but for #2.. do i do the same thing? or would i just apply the fundamental theorem of calculus without absolute values and without finding zeros?

A good example would be:
http://img400.imageshack.us/img400/6026/integralquestion7ht.gif

Is this done correctly? or would i still do any absolute value or splitting into regions with direction #2?

If I'm correct.. Direction #2 can have negative values or even 0, and direction #1 will always have positive values?

Please clarify all this someone.. thanks.

 

[verd]

I believe that the question is just asking you to set up two different integrals for the region -1 to 1... For this, you'd have to split it up into two pieces, as you did, and solve.

So direction one I think is asking you to just figure out and set up the integrals.

Direction two is asking you to solve, I believe. But what you did first was correct, I believe.

 

[Jacobpm64]

well, thing is.. it's not two different directions for the same problem.. i just used that example as.. an example lol... they have one whole section of about 10 problems with the 1st direction.. then they have another 10 problems asking to be solved using the 2nd direction..

it's not both for each problem

 

[d_leet]

well, thing is.. it's not two different directions for the same problem.. i just used that example as.. an example lol... they have one whole section of about 10 problems with the 1st direction.. then they have another 10 problems asking to be solved using the 2nd direction..

it's not both for each problem

I think you have the right idea, for the first part you need to split the integrals into parts that are either above or below the x-axis and then find the integral of each part, taking the opposite of the negative ones and then adding them. For the second part it would just be to evaluate the definite integral which could en up being positive, negative, or zero.

 

[Jacobpm64]

thanks for the clarification