Question

$$\begin{gathered} T\mathrm{he}\mathrm{functions}\mathrm{F},\mathrm{G}\mathrm{are}\mathrm{differentiable}\mathrm{als}oF\left(x\right)=G\left(x\right). \\ \mathrm{Where},\mathrm{C}\mathrm{is}\mathrm{any}\mathrm{constant}. \\ \mathrm{prove}\mathrm{that}G\left(x\right)=F\left(x\right)+C. \end{gathered}$$

Short answer

Proved

Step-by-step solution

Step 1. To Prove

$G\left(x\right)=F\left(x\right)+C.$

Step 2. Proving

$$\begin{gathered} Now, \\ F\text{'}\left(x\right)=G\left(x\right) \\ \int F\left(x\right)dx=\int G\left(x\right)dx \\ F\left(x\right)+C₁=G\left(x\right)+C₂ \\ G\left(x\right)=F\left(x\right)+(C₁-C₂) \\ G\left(x\right)=F\left(x\right)+C \\ \mathrm{Therefore},\mathrm{it}\mathrm{is}\mathrm{proved}\mathrm{tha}tG\left(x\right)=F\left(x\right)+C. \end{gathered}$$