The value of is
ok, let me explain
if we have the int{f(x)}dx = F(x), that is F(x) is the integral of f(x), then:
int{f(x)}dx from a going up to b = F(b) - F(a)
this is called the fundamental theorem of calculus.
so you do the integration as normal. when you end up with the final expression, you plug in the number at the top of the integral sign and subtract the expression when you plug in the number at the bottom of the integral sign.
so in this problem we saw the integral was -x^-1 (when finding the definate integral, that is integral between two given limits, we can forget about the + C since they'll cancel out anyway)
so if i let F(x) = -x^-1, then the integral between 6 and 10 is given by
F(10) - F(6)
and that's how i got -10^-1 - (-6^-1) = -10^-1 + 6^-1 = 1/15
so look up in your text book for the "fundamental theorem of calculus" it tells you how to deal with these problems
For this, we integrate (f(x) - g(x))dx from x = -5 to x = 6
INT {-5,6} [(.3x^2 + 3) - (x)] dx
INT {-5,6} [.3x^2 - x + 3] dx
[.1x^3 - .5x^2 + 3x] | {-5,6}
(.1(-5)^3 - .5(-5)^2 + 3(-5)) - (.1(6)^3 - .5(6)^2 + 3(6))
You should have no problem with the arithmatic.
Edit!
[.1x^3 - .5x^2 + 3x] | {-5,6} = (.1(6)^3 - .5(6)^2 + 3(6)) - (.1(-5)^3 - .5(-5)^2 + 3(-5))
Thanks, Jhevon
ecMathGeek is right! (though he did plug in the limits incorrectly). to find the area between two curves it is the integral of (higher curve - lower curve) between the desired limits
this one isn't so bad, look out for problems where the curves switch roles! that is the higher curve in one interval, is the lower curve in the other interval and you have to integrate across both intervals, in that case, you would split the integral in two with the respective limits