1 Answer | Add Yours
Let's take Newton-Leibniz formula: `int_a^bf(x)dx=F(b)-F(a)` where `F(x)=intf(x)dx`. So if we were to to calculate indefinite integral of e.g.`f(x)=x` we would get `F(x)=int xdx=x^2/2+C`. Let's now calculate definite integral: `int_a^b x dx=b^2/2+C-a^2/2-C=b^2/2-a^2/2` So we see that constants cancel each other so there is no need to write them at all. Also `C` can be any constant so we can choose it to be 0.
We’ve answered 288,106 questions. We can answer yours, too.Ask a question