Integration question?

i got this on my test today i was wondering if i did it right or not:


∫ f(x)dx=4 from [3,0], *where 3 is the upper and 0 is the lower


∫ f(x)dx=4 from [6,3]


∫ f(x)dx=5 from [6,2]


∫ f(x)dx=? from [2,0]





okay this is how i solved it:


∫ f(x)dx=4 from [3,0] =%26gt; F(3) - F(0) = 4 =%26gt; F(0) = F(3) - 4


∫ f(x)dx=4 from [6,3] =%26gt; F(6) - F(3) = 4


∫ f(x)dx=5 from [6,2] =%26gt; F(6) - F(2) = 5 =%26gt; F(2) = F(6) - 5


∫ f(x)dx=? from [2,0] =%26gt; F(2) - F(0) = [F(6) - 5] - [F(3) - 4] = F(6) - 5 - F(3) + 4 = F(6) - F(3) -1 = 4-1 = 3





if i do it the other way


∫ f(x)dx=4 from [3,0] =%26gt; F(3) - F(0) = 4 =%26gt; F(3) = 4, thus F(0)=0


∫ f(x)dx=4 from [6,3] =%26gt; F(6) - F(3) = 4 =%26gt; F(6)-4 = 4 =%26gt;F(6) =0


∫ f(x)dx=5 from [6,2] =%26gt; F(6) - F(2) = 5 =%26gt; 0-F(2) = 5 =%26gt;F(2)= -5


∫ f(x)dx=? from [2,0] =%26gt; F(2) - F(0) = -5 - 0 = -5





*I assume that F(0) =0 b/c fr what i understand if u find the anti-dev for anything function, let say f(x)=x+1, then F(x)=(x^2)/2 + x, thus F(0)=0


so which 1 is right? i put the 1st one on my test

Integration question?
F(0) doesn't always equal 0.


In your notation, I believe you are listing the upper number first and the lower number second. I will follow your notation.


Here's an example:


∫ sinx dx from [pi,0] = -cosx (from 0 to pi)


= -cos(pi) - (-cos0)


= -(-1) - (-1)


= 1 + 1 = 2


Here F(0) = -1





The first way is right.
Reply:WOW! Four votes for a no best answer. That's weird. Report It

Reply:Answer is 3....


Let Int=integral





Int(0....3)+Int(3....6)


=Int(0....6)





So 4+4=Int(0....6)=8





Int(0....2)+Int(2....6)


=Int(0....6)=8





Int(2...6)=5





So Int(0....2)=8-5=3