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\begin{center}

\Huge{Practice using the FTC}\\
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\normalsize
\noindent 
\begin{enumerate}
\item Complete the following statement of the Fundamental Theorem of Calculus: If $F'$ (the derivative of $F$) is continuous, then 

$$\int_a^b F'(x)\,dx=$$

\vskip .5in

\item Find the following exactly (no estimation) using the Fundamental Theorem of Calculus.
\begin{enumerate}
\item $\displaystyle{\int_{0}^5 \frac{1}{2}x^2 +\sin x \, dx}$
\vskip 2in
\item $\displaystyle{\int_{1}^{2.1} x^{-3} -e^x \,dx}$
\vskip 2in
\item$\displaystyle{\int_1^\pi \frac{1}{x} +2^x\ln 2\,dx}$
\end{enumerate}


\newpage
\item Look again at 2(a).  I'm guessing the antiderivative you choose to compute this integral was $F(x)=\frac{1}{6}x^3-\cos x$.  Consider the function $G(x)=\frac{1}{6}x^3-\cos x +10$.

\begin{enumerate}
\item Find $G'(X)$.
\vskip 1in
\item Use $G$ to compute  $\displaystyle{\int_{0}^5 \frac{1}{2}x^2 +\sin x \, dx}$ again.
\vskip 1.5in
\item Is your answer above the same as that you found in 2(a)?
\vskip .5in

\end{enumerate}

\noindent Read the text below Example 1 on page 309 to confirm your answer above and take note of the notation they use as I will adopt it today.

\item Find the following exactly (no estimation) using the Fundamental Theorem of Calculus and the new notation you just learned from page 309.
\begin{enumerate}
\item $\displaystyle{\int_0^1 6q^2+4\,dq}$
\vfill 
\item $\displaystyle{\int_1^{3.5} 2e^t \,dt}$ 
\vfill 

\end{enumerate}

\end{enumerate}
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