Theorem $3\mathrm{.}3\mathrm{.}14($Fundamental Theorem of Calculus$)$ If $f$ is continuous on $a,b,$ then $f$ has an antiderivative on $a,b.$ Moreover, if $F$ is any antiderivative of $f$ on $a,b,$ then

${\displaystyle \underset{a}{\overset{b}{}}}f(x)dx=F(b)-F(a)$

$f(x)=\{\begin{array}{ccc}{x}^{2}sin(\frac{1}{x})& if& x0\\ 0& if& x=0\end{array}$

is differentiable everywhere, but its derivative is discontinuous at $0.[$We found that $f^{\text{'}}(0)=0,$ and for $x0,f^{\text{'}}(x)=2xsin(\frac{1}{x})-cos(\frac{1}{x}),$ and that $\underset{x0}{lim}{f}^{\text{'}}(x)$ does not exist.$]$ Prove that $f^{\text{'}}$ is integrable on $0,\frac{2}{}$ and find ${\displaystyle \underset{0}{\overset{\frac{2}{}}{}}}{f}^{\text{'}}(x)dx.($Explain why Theorem $3\mathrm{.}3\mathrm{.}14$ cannot be used here.$)$