Answer
False
Work Step by Step
Consider $f(x)=\frac{1}{x}$
This function is continuous $[1,\infty)$ and is also decreasing on $[1,\infty)$ (because $f'(x)=\frac{1}{x^{2}}$ , which is always negative for any $x\ne 0$ ).
Also $\lim\limits_{n \to \infty}f(x)=0$
However,
$\int^{\infty}_{1}f(x)dx =\lim\limits_{x' \to \infty}\int^{x'}_{1}f(x)dx$
$=\lim\limits_{x' \to \infty}\int^{x'}_{1}\frac{1}{x}dx$
$=\lim\limits_{x' \to \infty} [lnx]^{x'}_{1}$
$=\lim\limits_{x' \to \infty} [lnx'-ln1]$
$=\lim\limits_{x' \to \infty} lnx'$
$=\infty $
Therefore, this integral is not convergent.
Hence, the given statement is false.
