Q. 77 - Calculus

Question

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

Step-by-Step Solution

Answer

Hence Proved

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

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