Proof of $\int f(x) \pm g(x)\,dx = \int f(x)\,dx \pm \int g(x)\, dx$.

Question

I want to prove that $\int f(x) \pm g(x)\,dx = \int f(x)\,dx \pm \int g(x)\, dx$.

I am familiar with the proof for this fact for derivatives using differentiation from first principles, but I am not too sure how to prove this property of integrals.

Any help would be greatly appreciated.

Answer 1

Suppose that $~F(x)~$ is an anti-derivative of $~f(x)~$ and $~G(x)~$ is an anti-derivative of $~g(x)~$. So by the definition of anti-derivative $~F'\left( x \right) = f\left( x \right)~$ and $~G'\left( x \right) = g\left( x \right)~$.

Also by the basic properties of derivatives $$~{\left( {F\left( x \right) \pm G\left( x \right)} \right)^\prime } = \,F'\left( x \right) \pm G'\left( x \right) = f\left( x \right) \pm g\left( x \right)~.$$

So $~\,F\left( x \right) + G\left( x \right)~$ is an anti-derivative of $~\,f\left( x \right) + g\left( x \right)~$ and $~\,F\left( x \right) - G\left( x \right)~$ is an anti-derivative of $~\,f\left( x \right) - g\left( x \right)~$.

Hence $$~\int\left[{{f\left( x \right) \pm g\left( x \right)}}\right] \,dx= F\left( x \right) \pm G\left( x \right) + c = \left(\int{{f\left( x \right)\,dx}}\right) \pm \left(\int{{g\left( x \right)\,dx}}\right)~.$$

Answer 2

Suppose $F$ is a primitive function of $f$ and $G$ primitive function of $g$, then: $$(F(x) + G(x))' = F'(x) + G'(x) =f(x) + g(x)$$ from this follows that $F + G$ is a primitive function of $f + g$.