This is the 6th project for Calc1 at Fitchburg State. Students are walked through the steps to justify the different pieces of the Fundamental Theorem of Calculus and make connections between the two parts.
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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 $F$.
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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}
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