Q. 79 - Calculus

Question

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

Step-by-Step Solution

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 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:

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