Question

$$\begin{gathered} \mathrm{The}\mathrm{objective}\mathrm{is}\mathrm{to}\mathrm{prove}\mathrm{that}\int \left(f\right(x)+g(x\left)\right)dx=\int f\left(x\right)dx+\int g\left(x\right)dx\mathrm{in}\mathrm{words}. \\ LetF\left(x\right)\mathrm{be}\mathrm{the}\mathrm{anti}-\mathrm{derivative}\mathrm{of}f\left(x\right)andG\left(x\right)\mathrm{be}\mathrm{the}\mathrm{anti}-\mathrm{derivative}\mathrm{o}fg\left(x\right). \end{gathered}$$

Short answer

Hence Proved

Step-by-step solution

Step1.  To prove

$\int \left(f\right(x)+g(x\left)\right)dx=\int f\left(x\right)dx+\int g\left(x\right)dx$

Step 2. Proving

$$\begin{gathered} \int \left(f\right(x)+g(x\left)\right)dx=\int f\left(x\right)dx+\int g\left(x\right)dx \\ \mathrm{The}\mathrm{sum}\mathrm{rule}\mathrm{of}\mathrm{derivative}\mathrm{says}\mathrm{that}. \\ \left(F\right(x)+G(x\left)\right)= \\ =\frac{d}{dx}F\left(x\right)+\frac{d}{dx}G\left(x\right) \\ =f\left(x\right)+g\left(x\right) \\ So, \\ f\left(x\right)dx+\int g\left(x\right)dx \\ =\left(F\right(x)+C)+\left(G\right(x)+C)[C=cons\tan t] \\ =\left(F\right(x)+G(x\left)\right)+C \\ =\int \left(f\right(x)+g(x)dx \\ \mathrm{Therefore},\int (f\left(x\right)+g\left(x\right))dx=\int f(x)dx+\int g(x)dx\mathrm{is}\mathrm{proved}. \end{gathered}$$