# 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 derivative and integral are almost inverse functions, so in turn, you are almost correct. For simple polynomials, one multiplies by the power and then removes 1 from the power, and the other adds 1 to the power and divide by the new power.

For more complex functions, you can consider it visually, or even compare it to physics. If you have a line (velocity), the gradient is the acceleration. If you derive this line to get the gradient, you know have the acceleration function. Now, if you have a flat line with no gradient (acceleration), and you integrate it, you will be left with a line with gradient for the velocity function.

This is because acceleration represents rate of change of distance relative to time, just like how gradient represents rate of change of y relative to x.

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\$