Integral of a deriative.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? Tha in advance

Question

Integral of a deriative.I&#039;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? Tha in advance

alkaholikd9 · Accepted Answer

Since ∑axf′(t)dt=f(x)−f(a) (1)the short answer is that the integral of the derivative is the original function, up to a constant.Of course, (1) isn&#039;t true without restrictions. But if f′ is continuous, then, yes, (1) holds.

Wendy Boykin · Accepted Answer

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)=2x2+3x+6, you&#039;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.

nick1337 · Accepted Answer

The integral of the derivative isn&#039;t always equal to the original function.example : let f be a function asf(x)=2x+2so we have f′(x)=2If you integrate f&#039;, you&#039;ll end up with F(x)=2x+cwith c a real constant. So you&#039;ll have your initial function only if c=2

Integral of a deriative. I've been learning the fundamental theore

Answered question

Answer & Explanation

New Questions in Calculus 1