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{\scshape Math 2300} \hfill {\scshape \Large Project \#6} \hfill {\scshape Fall 2017}
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{\bf Consider the graph of $y = 2t -2$.}
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\begin{enumerate}
\item Sketch the region below this line, above the $t$-axis, and between the vertical lines $t=1$ and $t = 4$.
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\item Use geometry to find the area of the region.
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\item Now sketch the region below the line $y=2t -2$, above the $t$-axis, and between the lines $t=1$ and $t=x$ for some $x > 1$.
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\item Use geometry to find the area of this region as a function of $x$. Call this area, your function, $A(x)$.
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\item Take the derivative of the area function $A(x)$.
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\end{enumerate}
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{\bf Now, consider the function $y = 3 + t^2$.}
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\begin{enumerate}
\item For some $x>1$, sketch the region that the function $\displaystyle{A(x) = \int_{-1}^x (3 + t^2)\,dt}$ represents the area of.
\item Use the fact that $\displaystyle{\int_a^b u^2\,du = \frac{b^3-a^3}{3}}$ and $\displaystyle{\int_a^bc\,du = c(b-a)}$, and the rules of combining definite integrals to find an expression for $A(x)$ and simplify that expression.
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\item Compute $A'(x)$.
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\item For a small positive number $h$, sketch the region whose area is represented by \\$A(x + h) - A(x)$.
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\item Use your picture, and maybe a rectangle, to explain why $\displaystyle{\frac{A(x+h) - A(x)}{h} \approx 3 + x^2}$.
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\item Based on part (e), give both an intuitive reason and a logical reason using the limit definition of the derivative for why your answer in (c) makes sense.
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\end{enumerate}
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{\bf Suppose that $f$ is a continuous function. Define a new function $g$ by $\displaystyle{g(x) = \int_a^x f(t) \, dt}$, where $a$ is a real number and $x>a$. Based on your above work take a guess at what $g'(x)$ is.}
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\begin{center} THIS IS PART ONE OF THE FUNDAMENTAL THEOREM OF CALCULUS! \end{center}
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{\bf Now you've seen that for a continuous function $f$, that if $\displaystyle{\int_a^xf(t)\,dt}$, then $g'(x) = f(x)$. Remember that the lower limit, $a$, is a number, while the upper limit, $x$, is the variable which we are taking the derivative with respect to. Use this theorem to find $g'(x)$ in the following. Practice applying this new differentiation rule and combining it with previous rules as well as the properties of definite integrals that we learned in section 5.2}
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\begin{enumerate}
\item $\displaystyle{g(x) = \int_\pi^x \sin (2t)\,dt}$.
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\item $\displaystyle{g(x) = \int_x^7 t^3 - \frac{1}{t}\,dt}$.
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\item $\displaystyle{g(x) = \int_0^{x^4} \sec t\,dt}$.
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\item $\displaystyle{g(x) = \int_{2t}^{3t} \frac{3t + 1}{t^2 + 1}\,dt}$.
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\end{enumerate}
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{\bf Let $\displaystyle{g(x) = \int_a^x f(t)\,dt}$, for a continuous function $f$ on the interval $[a,b]$. Let's also suppose that $F(x) = \int f(x)\,dx$.}
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\begin{enumerate}
\item What are $g'(x)$ and $F'(x)$? What does that tell you about the difference (like subtraction) between $g(x)$ and $F(x)$?
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\item Write an equation that shows what you concluded in part (a).
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\item Compute $g(a)$ in two ways: using the definition of $g$, and also using your formula from part (b).
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\item These two methods should give you the same solution, setting these two answers equal allows you to solve for a constant. Do it.
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\item Rewrite the definite integral $\displaystyle{\int_a^b f(t)\,dt}$ in terms of the function $g$.
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\item Use the formula, finished in part (d) when you solved for the constant, to solve the definite integral in terms of $F$.
\end{enumerate}
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\begin{center} THIS IS PART TWO OF THE FUNDAMENTAL THEOREM OF CALCULUS! \end{center}
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{\bf This is the portion of the FUNdamental Theorem that we will use most often. Now, go forth and integrate!}
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\begin{enumerate}
\item $\ds{\int_0^1\! x^{4/5}\,dx}$
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\item $\ds{\int_1^e \! \frac{1}{x}\,dx}$
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\item $\ds{\int_{-1}^1\!e^{u+2}\,du}$
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\item $\ds{\int_{-\pi/2}^{\pi/2} \cos t\,dt}$
\end{enumerate}
\end{document}
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