utf22-003
Find the average value of the function on the interval, then
find the ``c'' guaranteed by the Mean Value Theorem for Integrals. Graphically
show the areas which the theorem implies are equal for part (b).
$$a) f( x) = \csc^2 x \text{on} [\frac{\pi }{ 4},{\frac{3\pi }{ 4}}]
b) f( x) = x(x-1)^5 \text{on} [1,2] $$
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utf22-004
Calculate the following derivatives using the chain rule and
the fundamental theorem of calculus.
\begin{alignat*}{3}
&a) \frac{ds}{dx}{\int_0^x {\frac{\text{dt} }{ 1+t^2}}} &
&b) \frac{ds}{dx}{\int_0^{x^2}{\frac{\text{dt} }{ 1+t^2}}} &
&c) \frac{ds}{dx}{\int_{{-x^2}}^{x^2}{\frac{\text{dt} }{ 1+t^2}}} \\
&d) \frac{d^2 }{ dx^2}[{\int_0^x {\frac{\text{dt} }{ 1+t^2}}}] &
&e) \frac{ds}{dx}{\int_{1}^{{{\tan x}}}{{t^{10}\cos t}} dt} &
&f) \frac{ds}{dx}{\int_{{{x^3}}}^{{{x^5 +1}}}{\frac{1}{ t}} dt}
\end{alignat*}
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utf22-006
We are given a differentiable, odd function $ f( x)$ defined
on $ [-3,3] $ which has zeros at $x=-2,0, \text{and} 2$ (nowhere else)
and critical points at $x=-1$ and $x=1$ (nowhere else). Also, we know that
$ f( {-1})=1.$ Define a new function $F$ on $ [-3,3] $ by the formula
$$ F(x) = \int_{{{-2}}}^{x}{ f( t)} dt$$
Sketch a rough graph of $ f( x).$
Find the value of $ F({-2}), F({2}), $ and an upper and lower
bound on $ F(0).$
Find the critical points and inflection points of $ F(x)$ on
$ [-3,3].$
Sketch a rough graph of $ F(x)$ on $ [-3,3].$
Interpret the points found in (c) in terms of the graphs of $ f( x)$
and $ F(x).$
The region under the graph of $y=-2x +4$ on $ [-2,1] $ is to be divided
into two parts of equal area by a vertical line. Where should the line
be drawn?
Where would you draw a horizontal line to divide the region in part (a)
into two parts of equal area?