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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

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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 $.

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    $\begingroup$ 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? $\endgroup$ – Nap D. Lover Aug 9 '17 at 11:09
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    $\begingroup$ @LoveTooNap29 I did. Thanks for pointing that out! $\endgroup$ – LloydTao Aug 9 '17 at 11:13
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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.

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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 $c=2$

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