Chapter 6.3, Problem 53E

To determine

To show: $F(b)-F(a)=G(b)-G(a)$

Explanation of Solution

Proof:

An anti-derivative of $F\text{'}\left(x\right)$ is $F\left(x\right)$ and an anti-derivative of $G\text{'}\left(x\right)$ is $G\left(x\right).$

Since,

  ${{\displaystyle \int }}_a^{x}f(t)dt=F(x)-F(a)$

Where F is any anti-derivative of f

So,

  ${{\displaystyle \int }}_a^b{F}^{\prime }(x)dx=F(b)-F(a)$

  ${{\displaystyle \int }}_a^b{G}^{\prime }(x)dx=G(b)-G(a)$

Since,

  $F^{\prime }(x)=G^{\prime }(x)$

Take integral on both sides-

  ${{\displaystyle \int }}_a^b{F}^{\prime }(x)dx={{\displaystyle \int }}_a^b{G}^{\prime }(x)dx$

So,

  $F(b)-F(a)=G(b)-G(a)$