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\centerline{MATH 136 -- Calculus 2}
\centerline{Practice Day on Consequences of FTC, ``Part I''}
\centerline{January 31, 2020}
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\noindent
{\it Background}
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Recall that the ``FTC, part I'' says:  If $f$ is continuous on $[a,b]$ and
$F$ is any function satisfying $\frac{d}{dx} F(x) = F'(x) = f(x)$ for all $x$ in $[a,b]$ (called
an \emph{antiderivative} of $f$), then
$$\int_a^b f(x)\ dx = F(b) - F(a).$$
We will use the \emph{shortcut notation}:  $\left. F(x)\right|_a^b$ for this difference of
values at the limits of integration:
$$\left. F(x)\right|_a^b = F(b) - F(a).$$
Note that you always put the value $F(b)$ at the \emph{upper limit} first and then subtract
the value at the lower limit, $F(a)$.
This means that we have a ``shortcut method'' for computing $\int_a^b f(x)\ dx$
as long as we have or can find a suitable \emph{antiderivative} $F$ -- a function
with $F'(x) = f(x)$ for the function $f(x)$ we are integrating.
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\noindent
{\it Questions}

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\begin{enumerate}
\item[(1)] Verify that 
$$\frac{d}{dx}(e^{x^2}) = 2x e^{x^2}.$$
\item[] {\it Answer:}  This follows from differentiation using the chain rule.
\item[(2)] Use part (1) to evaluate:
$$\int_0^1 2x e^{x^2}\ dx$$
\item[] {\it Answer:}  Using (1), 
$$\int_0^1 2x e^{x^2}\ dx = \left. e^{x^2} \right|_0^1 = e - 1 \doteq 1.71828.$$
\item[(3)] Verify that 
$$\frac{d}{dx}(\ln(x^2+5x+6)) = \frac{2x+5}{x^2 + 5x + 6}.$$
\item{} {\it Answer:}  Differentiate using the chain rule and the derivative
formula for $\ln(x)$.
\item[(4)]  Use part (4) to evaluate:
$$\int_1^2 \frac{2x + 5}{x^2 + 5x + 6}\ dx.$$
\item[] {\it Answer:}  Using (3), 
$$\int_1^2 \frac{2x + 5}{x^2 + 5x + 6}\ dx = \left. \ln(x^2 + 5x + 6)\right|_1^2 = \ln(20) - \ln(12) = \ln(5/3) \doteq .5108$$
\item[(5)]  Verify that 
$$\frac{d}{dx}\left(\frac{1}{2}\left(1 + \sin(x)\cos(x)\right)\right) = \cos^2(x)$$
(You'll need a trig identity here -- ask if you don't recall identities using
$\sin^2(x)$ and $\cos^2(x)$!)
\item[] {\it Answer:}  Here's one way (there are others too):  By the product rule and the identity 
$\sin^2(x) + \cos^2(x) = 1$, we have:
\begin{align*} 
\frac{d}{dx}\left(\frac{1}{2}\left(x + \sin(x)\cos(x)\right)\right) &=
\frac{1}{2}(1-\sin^2(x) + \cos^2(x))\\
&= \frac{1}{2}(2\cos^2(x))\\
&= \cos^2(x)
\end{align*}
\item[(6)]  Use part (5) to evaluate:
$$\int_0^{\pi/2} \cos^2(x)\ dx.$$
\item[] {\it Answer:}
$$\int_0^{\pi/2} \cos^2(x)\ dx = \left.\frac{1}{2}\left(x + \sin(x)\cos(x)\right)\right|_0^{\pi/2} = \pi/4$$
(recall $\cos(\pi/2) = 0$ and $\sin(0)= 0$ -- this means the only nonzero term comes from the $x$ at
the upper limit of integration).
\end{enumerate}

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