*Question 1*
(a) We have
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| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------- |
| $F(x)=\left( 1 -\frac{1}{2}x + \frac{1}{12}x^{2} \right)e^{x}
lt;br> | $F(0)=1$ |
| $F'(x) = \left( -\frac{1}{2} + \frac{1}{6}x + 1 - \frac{1}{2} x + \frac{1}{12}x^{2} \right)e^{x} = \left( \frac{1}{2} -\frac{1}{3} x + \frac{1}{12}x^{2} \right)e^{x}lt;br> | $F'(0)=\frac{1}{2}$ |
| $F''(x)=\left( -\frac{1}{3} + \frac{1}{6}x + \frac{1}{2} -\frac{1}{3}x +\frac{1}{12} x^{2} \right)e^{x} =\left( \frac{1}{6} -\frac{1}{6} x + \frac{1}{12}x^{2} \right)e^{x}$ | $F''(0)=\frac{1}{6}$ |
| $F'''(x) = \frac{1}{12} x^{2} e^{x}$ | $F'''(t)=\frac{1}{12} t^{2}e^t$ |
So by Taylor's theorem with $n=3,$ $\begin{align}
F(x) &= F(0) + F'(0)\,x + \frac{F''(0)}{2} x^{2} + \frac{F'''(t)}{6} x^{3} \\
&= 1 + \frac{1}{2}x + \frac{1}{12}x^{2} + \frac{1}{72}t^{2} e^{t} x^{3} \\
\end{align} $for some $t$ between $0$ and $x.$ For such $t$ and $x\geq 0$ we have $\frac{1}{72} t^{2}e^{t}x^{3}\geq 0$ so $F(x) \geq 1 + \frac{1}{2}x + \frac{1}{12}x^{2}$
(b) Note that $\frac{1}{12}x^{2} - \frac{1}{2}x + 1 = \frac{1}{12} (x - 3)^{2} + \frac{1}{4}\geq \frac{1}{4} >0.$
(c) From (a) for $x\geq 0$, $\left( 1-\frac{1}{2}x + \frac{1}{12}x^{2} \right)e^{x} \geq 1+ \frac{1}{2}x + \frac{1}{12}x^{2}$Dividing both sides by $1-\frac{1}{2}x+\frac{1}{12}x^{2}> 0$ gives $e^{x} \geq \frac{ 1+\frac{1}{2}x+\frac{1}{12} x^{2} }{1-\frac{1}{2}x +\frac{1}{12}x^{2}}$so $e= e^{1}\geq \frac{19}{7}.$
*Question 2*
Since $f,g$ are integrable they are also bounded so let $|f|,|g|\leq M$ for some $M\geq 0.$ Set $\varepsilon>0.$ Choose a partition $P$ of $[a,b]$ so that $U(f,P)< \overline{\int} f + \varepsilon $Suppose $f,g$ differ at $n\in \mathbb{N}$ points. Let $Q$ be a refinement of $P$ such that the length of each subinterval is less than $\delta:= \frac{\varepsilon}{2Mn}.$ It follows that $|U(f,Q)-U(g,Q)|<\delta \cdot 2 M n = \varepsilon$since $f$ and $g$ differ in at most $n$ intervals by at most $2M$ and are equal in every other interval.
Now $ \overline{\int} g \leq U(g,Q) < U(f,Q ) + \varepsilon \leq U(f,P)+\varepsilon < \overline{\int} f + 2\varepsilon$Hence $\overline{\int}g - \overline{\int}f < 2\varepsilon \implies \overline{\int} f = \overline{\int} g$ since $\varepsilon$ is arbitrary.
Similarly $\underline{\int} f = \underline{\int}g.$
*Question 3*
Let $P_{n}$ be the partition of $[0,1]$ into $n$ equal intervals. We've shown that for any continuous function $f:[0,1]\to \mathbb{R},$ we have $U(f,P_{n}) \to \int_{0}^{1} f(x) \, dx$ as $n\to \infty.$
Now define $f(x)=x^{p}.$ Then the maximum of $f$ on the $k$th part of $P_{n}$ is $f\left( \frac{k}{n} \right)=\frac{k^{p}}{n^{p}}.$ So the upper sum is $U(f,P_{n})= \frac{1}{n^{p+1}} \sum_{k=1}^{n} k^{p+1}$which converges to $\int_{0}^{1} x^{p} \, dx = \left[ \frac{x^{p+1}}{p+1} \right]_{0}^{1} = \frac{1}{p+1} $as $n\to \infty.$