Let $fbe$ continuous and $gbein{C}^1[a,b].$ Prove that ${\displaystyle \underset{a}{\overset{g(x)}{}}}f(t)dtisin{C}^1[a,b]$ and $\frac{d}{dx}{\displaystyle \underset{a}{\overset{g(x)}{}}}f(t)dt=$

$f(g(x))g^{\text{'}}(x).$

Proof Let $F(x)={\displaystyle \underset{a}{\overset{x}{}}}f(t)dt$ for any $x.$ This implies $F(g(x))={\displaystyle \underset{a}{\overset{g(x)}{}}}f(t)dt.By$ the chain

rule,

$[F(g(x)){]}^{\text{'}}=$ubrace$(F^{\text{'}}(g(x))ubrace)g^{\text{'}}(x)$

$=f(g(x))g^{\text{'}}(x)FTC$ applied $to$ the under braced term