Question

Prove that if two functions F and G differ by a constant, then ${\left[F\right(x\left)\right]}_a^b={\left[G\right(x\left)\right]}_a^b$.

Short answer

Ans:

$$\begin{gathered} \left[\mathrm{G}\right(\mathrm{x}){]}^{*} \\ =[\mathrm{F}\left(\mathrm{x}\right)+\mathrm{C}{]}_0^{\mathrm{k}} \\ =\left(\mathrm{F}\right(\mathrm{b})+\mathrm{C})-\left(\mathrm{F}\right(\mathrm{a})+\mathrm{C}) \\ =\mathrm{F}\left(\mathrm{b}\right)-\mathrm{F}\left(\mathrm{a}\right) \\ =\left[\mathrm{G}\right(\mathrm{x}){]}_4^{\mathrm{b}} \end{gathered}$$

Step-by-step solution

Step 1. Given Information: 

Functions F and G differ by a constant

${\left[F\right(x\left)\right]}_a^b={\left[G\right(x\left)\right]}_a^b$

Step 2. Prove:

$$\begin{gathered} \mathrm{Now},\mathrm{it}\mathrm{is}\mathrm{clear}\mathrm{that}\mathrm{G}\left(\mathrm{x}\right)-F\left(x\right)=C\Rightarrow G\left(x\right)=F\left(x\right)+C. \\ \left[\mathrm{G}\right(\mathrm{x}){]}^{*} \\ =[\mathrm{F}\left(\mathrm{x}\right)+\mathrm{C}{]}_0^{\mathrm{k}} \\ =\left(\mathrm{F}\right(\mathrm{b})+\mathrm{C})-\left(\mathrm{F}\right(\mathrm{a})+\mathrm{C}) \\ =\mathrm{F}\left(\mathrm{b}\right)-\mathrm{F}\left(\mathrm{a}\right) \\ =\left[\mathrm{G}\right(\mathrm{x}){]}_4^{\mathrm{b}} \\ \therefore henceproved{\left[F\right(x\left)\right]}_a^b={\left[G\right(x\left)\right]}_a^b \end{gathered}$$