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\centerline{MATH 136 -- Calculus 2}
\centerline{Practice Day on Consequences of FTC, ``Part II''}
\centerline{January 29, 2020}
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\noindent
{\it Background}
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Recall that the ``FTC, part II'' says:  If $f$ is continuous on $[a,b]$, then
$$\frac{d}{dx} \int_a^x f(t)\ dt = f(x)$$
for all $x$ in $[a,b]$.  (Note: at the endpoints $x = a$ and $x = b$, we may
be talking only about ``one-sided derivatives'' if $f$ is not defined for $x < a$
or $x > b$.)  

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\noindent
{\it Questions}

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\begin{enumerate}
\item[(0)]  Look at the function $f(t)$ plotted on the back.  What is 
$\frac{d}{dx} \int_0^x f(t)\ dt$ at $x = 3$? 
\item[] {\it Answer:}  It is $f(3) = 2$.
\item[(1)]  What is 
$$\frac{d}{dx} \int_3^x \sqrt[3]{t^5 + 3t + 1}\ dt?$$
\item[] {\it Answer:}  This is $\sqrt[3]{x^5 + 3x + 1}$.
\item[(2)]  Same question for 
$$\frac{d}{dx} \int_0^x \frac{\tan(t)}{t^3 + 2}\ dt$$
\item[] {\it Answer:}  This is $\frac{\tan(x)}{x^3 + 2}$.
\item[(3)]  What is 
$$\frac{d}{dx} \int_x^4 e^{4t^2 + 7}\ dt?$$
\item[] {\it Answer:}  Since the $x$ is in the lower limit, we need to reverse
the limits of integration, which introduces a negative sign, then apply ``FTC, part II''
\begin{align*}
&= - \frac{d}{dx} \int_4^x e^{4t^2 + 7}\ dt\\
&= - e^{4x^2 + 7}.
\end{align*} 
the derivative is $-e^{4x^2 + 7}$.
\item[(4)]  What about
$$\frac{d}{dx} \int_0^{x^2} \sqrt{t^2 + 1}\ dt?$$
\item[]  {\it Answer:}  Now we need the FTC, part II, and the chain rule: To see this note that we
are looking at $F(x^2)$ where $F(x) = \int_0^x \sqrt{t^2 + 1}\ dt$.  Then by the chain rule,
$\frac{d}{dx} F(x^2) = F'(x^2)\cdot 2x$.  The FTC, part II gives $F'(x^2) = \sqrt{x^4 + 1}$
and $2x$ is the ``derivative of the inside $x^2$.'' So the final answer is $2x\sqrt{x^4 + 1}$.  
\item[(5)]  What is 
$$\frac{d}{dx} \int_{-x^3}^{x^4} \frac{t}{t + 1}\ dt$$
(You'll need to use the interval union property, the interchange of limits
property and the chain rule here!)
\item[] {\it Answer:} Using the hints:
\begin{align*} 
\frac{d}{dx} \int_{-x^3}^{x^4} \frac{t}{t + 1}\ dt &= 
\frac{d}{dx} \left(\int_{-x^3}^{0} \frac{t}{t + 1}\ dt + \int_{0}^{x^4} \frac{t}{t + 1}\ dt\right)\\
&= \frac{d}{dx} \left(-\int_0^{-x^3} \frac{t}{t + 1}\ dt + \int_{0}^{x^4} \frac{t}{t + 1}\ dt\right)\\
&=\frac{-x^3}{-x^3+1}\cdot (-3x^2) + \frac{x^4}{x^4 + 1} \cdot 4x^3.
\end{align*}
\item[(6)]  Finally, suppose someone asked you for a function whose derivative
was $f(x) = \displaystyle{\frac{x^2}{x^4 + x + 1}}$.  How could you ``cook one up'' using
what we have talked about?  (In case you are worried about possible 
discontinuities from vertical asymptotes, you don't need to: this function $f$ is
continuous for all real $x$ -- the denominator is never zero.)
\item[] {\it Answer:}  The FTC, part II says that 
$$F(x) = \int_{0}^x \frac{t^2}{t^4 + t + 1}\ dt$$
is one such function.  Starting from any lower limit of integration $a$ rather than
zero gives others.  
\end{enumerate}
\begin{figure}
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\caption{The region}
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