Q. 64 - Calculus

Question

Suppose $f$ is continuous on all of $R$. Prove that for all real numbers $a$ and $b$, the functions  $A\left(x\right)={\int }_a^{x}f\left(t\right)dt$ and  $B\left(x\right)={\int }_b^{x}f\left(t\right)dt$ differ by a constant. Interpret this constant graphically.

Step-by-Step Solution

Answer

For all real numbers $a$ and $b$, the functions $A\left(x\right)={\int }_a^{x}f\left(t\right)dt$ and $B\left(x\right)={\int }_b^{x}f\left(t\right)dt$ differ by a constant.

Step 1. Given information

We have to prove that for all real numbers $a$ and $b$, the functions  $A\left(x\right)={\int }_a^{x}f\left(t\right)dt$ and $B\left(x\right)={\int }_b^{x}f\left(t\right)dt$ differ by a constant.

Step 2. Proof of the given question.

Since $f$ is continuous on all $R$,

$$\begin{gathered} A\left(x\right)-B\left(x\right) \\ ={\int }_a^{x}f\left(t\right)dt-\left({\int }_b^{x}f\left(t\right)dt\right) \\ ={\int }_a^{x}f\left(t\right)dt+{\int }_x^{b}f\left(t\right)dt \\ ={\int }_a^{b}f\left(t\right)dt \\ =F\left(b\right)-F\left(a\right) \end{gathered}$$

$F\left(b\right)-F\left(a\right)$ is a constant.

Therefore, the functions $A\left(x\right)={\int }_a^{x}f\left(t\right)dt$ and $B\left(x\right)={\int }_b^{x}f\left(t\right)dt$ differ by a constant.