Sorry if integral is not the correct english term.. I'm working with integrals and came across something that I haven't done before, thanks math book for the great explanation...

Rough translation:
Decide the area under the curve y = 4-x^2 in the interval: 0 ≤ x ≤ 1.

I have no idea how to even set up the problem.. Normally I have something along the lines of;

b
∫ f(x)[ ]
a

Any help is of course greatly appreciated, I apologize for not being able to give fg for help. ;>

My thoughts are that 0 represents a and 1 represents B. Then just take the "anti derivitive" and.. well, do as you normally do. However when I did that I got a completely different answer than the one given in the answer key.

This post was edited by Deadreborn on Aug 20 2013 12:10pm

uhh can you double check that interval to make sure it is correct?

This post was edited by Azrad on Aug 20 2013 12:04pm

My mistake, bloody alt buttons. It's 0 ≤ x ≤ 1. Will edit OP.


I've set it up and exucted in this manner;

1
∫ f(4-x^2)[4x-(x^3)/3 ] = (4*1-(1^3)/3) - ( 4*0-(0^3)/3) = (1) - (0) = 1 which is completely incorrect... According to my answer key it should be 3*(2/3) = 3.67
0

I feel like I'm missing something stupid and small somewhere.

This post was edited by Deadreborn on Aug 20 2013 12:34pm

yes you are missing something. you forgot how to subtract XDDD

check your final arithmetic... remember what 1 to the power of 3 is LOL XDD

(4*1-(1^3)/3) - ( 4*0-(0^3)/3) = 4 - 1/3 - 0 - 0 = 4-1/3 = 8/3

the division by 3 does not include 4 is the first parenthesis

you sure about 3*2/3 = 3.67 ? cause it should be 3*2/3 = 2 .. i guess you meant 3 + 2/3, which is indeed equal to 3.67, which is also equal to 8/3

Oh wow, I'm a silly fool aren't I. Something this simple, I should have noticed. I guess that's what 10hours of math every day for the past week gets you. Thanks so much for that, I was pulling my hair out. Time for more coffee.....

Silly mistakes of mine, *sigh*