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\problemgrouptitle{The Definite Integral}

\item Consider the following graphically defined function $f(x)$:

~~~\resizebox{.4\textwidth}{!}{%
\includegraphics{graphics/the-definite-integral1}%
}
\begin{enumerate}
\item Using the interpretation of the definite integral in terms of net signed area between the function and the $x$-axis, find
\begin{enumerate}
\item $\displaystyle \int_{-5}^0f(x)\,dx$
\item $\displaystyle \int_{-2}^5f(x)\,dx$
\end{enumerate}
\item If we define the function $\displaystyle g(t)=\int_{-6}^t f(x)\,dx$, on what intervals is
\begin{enumerate}
\item $g(t)$ increasing?
\item $g(t)$ decreasing?
\end{enumerate}
\end{enumerate}
\answer{\begin{enumerate}
\item \begin{enumerate}
\item $-6$
\item $4$
\end{enumerate}
\item \begin{enumerate}
\item Increasing on $(-6,-5)\cup(0,5)$
\item Decreasing on $(-5,0)\cup(5,6)$
\end{enumerate}
\end{enumerate}}

\begin{problemblock}[4]{Use the interpretation of the definite integral as the net signed area to find:}

\addtoanswerkey{\begin{multicols}{2}}

\item $\displaystyle\int_0^2 3x\d{x}$
\answer{$\displaystyle 6$}

\item $\displaystyle\int_{-1}^4 6\d{x}$
\answer{$\displaystyle 30$}

\item $\displaystyle\int_0^3(2x-4)\d{x}$
\answer{$\displaystyle -3$}

\item $\displaystyle\int_{-4}^3(-x)\d{x}$
\answer{$\displaystyle\frac{7}{2}$}

\end{problemblock}

\addtoanswerkey{\end{multicols}}

\item Find $\displaystyle\int_{-r}^0\sqrt{r^2-x^2}\d{x}$ by interpreting the integral as an area.  Here $r>0$ is a positive constant.
\answer{Notice that the area can be viewed as the quarter of a circle:

\resizebox{0.4\textwidth}{!}{%
\includegraphics{graphics/the-definite-integral2}%
}

Then the integral is $\displaystyle\frac{1}{4}\left(\pi r^2\right)$.  (Since the area is above the $x$-axis the integral is positive.)}

\item Simplify the following to a single definite integral using the properties of the definite integral.
\[\int_{-1}^7\!f(x)\,dx+\int_3^{-1}\!f(x)\,dx+\int_7^9\!f(x)\,dx\]
\answer{$\displaystyle\int_3^9\!f(x)\,dx$}

\item Use properties of the definite integral to verify the following without evaluating the integral.
  \[2\le\int_{-1}^1 (2+\sin x)\,dx\le 6\]
\answer{Since $-1\le\sin x\le 1$ it follows (adding 2 to these inequalities) that $m=1\le 2+\sin x\le 3=M$.  Since $b=1$ and $a=-1$ we have that $m(b-a)=1(2)=2$ and $M(b-a)=3(2)=6$.}

\begin{problemblock}[2]{Use Riemann sums with right endpoint evaluation to evaluate the following definite integrals.}

\item $\displaystyle\int_0^3 \left(x^3+1\right)dx$
\answer{$S_n=\displaystyle\frac{3}{n}\left(\frac{27n^4+54n^3+27n^2}{4n^3}+n\right)= \frac{93n^2+162n+81}{4n^2}$\newline
and so $\displaystyle\int_0^3\left(x^3+1\right)dx=\lim_{n\to\infty} S_n=\frac{93}{4}$.}

\item $\displaystyle\int_0^b x^2\d{x}$ where $b>0$ is constant
\answer{$S_n=\displaystyle b^3\left(\frac{1}{3}+\frac{1}{2n}+\frac{1}{6n^2}\right)$ and so $\displaystyle\int_0^b x^2\d{x}=\lim_{n\to\infty} S_n=\frac{b^3}{3}$.}

\end{problemblock}

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