Why is there always constant of integration when you evaluate an indefinite integral?

When you are asked to integrate a function f(x), you are really being asked the question: "what function F(x) exists such that when you take its derivative, you are left with f(x)?"

Let us first consider differentiation.

Let F(x)=x2

We know the derivative of this is f(x)=2x but what if F(x)=x2+5? 

It turns out the derivative of this is also f(x)=2x. That is because the derivative of 5 is 0 and so that disappears from the derivative.

In fact the derivative of any constant is 0 so the derivative of F(x)=x2+C (where C is any real number) is f(x)=2x

So now let us talk about integration.

We know that integrating f(x)=2x gives F(x)=x2 because when you differentiate F(x) you are left with f(x). But this is also true for F(x)=x2+5  or in fact for F(x)=x2+C (where C is any real number).

And therefore the fact that the derivative of any constant is 0 is the origin of the constant of integration.

AS

Related Maths A Level answers

All answers ▸

At x=3, is the polynomial y= (4/3)x^3 -6x^2 + 11 at a maxima or minima?


Find the integral of xcosx(dx)


Find the equation of the tangent to the curve x^3+yx^2=1 at the point (1,0).


Show that the line y = x - 7 does not meet the circle (x + 2)^2 + y^2 = 33.