Integral of a derivative.

I've been learning the fundamental theorem of calculus. So, I can intuitively grasp that the derivative of the integral of a given function brings you back to that function. Is this also the case with the integral of the derivative? And if so, can you please give a intuition for why this is true? Thanks in advance

The only main difference is that integrating leaves you with an unknown constant $C$. You may notice that if you differentiate $f(x) = 2x^2 + 3x + 6$, you're left with $f'(x) = 4x + 3$, and the $6$ has absolutely no effect on the final answer. This is because, no matter where the line/curve is located in the y-axis, the gradient for the x co-ordinate remains the same. You require a co-ordinate from the original function in order to calculate $C$.
• In the final paragraph you write "if you integrate $f(x)=..." followed by "[then] you're left with f'(x)=...", did you mean to write "differentiate" instead? – Nap D. Lover Aug 9 '17 at 11:09 • @LoveTooNap29 I did. Thanks for pointing that out! – LloydTao Aug 9 '17 at 11:13 Since$$\int_a^xf'(t)\,\mathrm dt=f(x)-f(a),\tag{1}$$the short answer is that the integral of the derivative is the original function, up to a constant. Of course,$(1)$isn't true without restrictions. But if$f'$is continuous, then, yes,$(1)$holds. The integral of the derivative isn't always equal to the original function. example : let$f$be a function as $$f(x) = 2x+2$$ so we have $$f'(x)= 2$$ If you integrate$f'$, you'll end up with $$F(x) = 2x + c$$ with c a real constant. So you'll have your initial function only if$c=2\$