2. Let f be the function shown below, with domain the closed interval [0, 6]. The graph of f is shown below, composed of two semicircles. Let h(x) = ∫ f (t)dt from x=0 to x=(2x-1). ( I don't think the image will load... it is a semicircle above the x-axis with r=2 and center (2,0) connected to a semicircle below the x-axis with r=1 and center (5,0).) A. Determine the domain of h(x). For this, I solved 1/2<=x<=7/2 B. Find h′(5/2) ... this is where I am lost, I don't know how to find the derivative of this unknown function. C. At what x is h(x) a maximum? Show all the analysis tha

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2. Let f be the function shown below, with domain the closed interval [0, 6]. The graph of f is shown below, composed of two semicircles. Let h(x) = ∫ f (t)dt from x=0 to x=(2x-1). ( I don't think the image will load... it is a semicircle above the x-axis with r=2 and center (2,0) connected to a semicircle below the x-axis with r=1 and center (5,0).) A. Determine the domain of h(x). For this, I solved 1/2<=x<=7/2 B. Find h′(5/2) ... this is where I am lost, I don't know how to find the derivative of this unknown function. C. At what x is h(x) a maximum? Show all the analysis tha

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is your function like this: |dw:1435084959591:dw|
Yes
we can write this: \[\Large f\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {\sqrt {4 - {{\left( {t - 2} \right)}^2}} ,\quad 0 \leqslant t \leqslant 4} \\ { - \sqrt {1 - {{\left( {t - 5} \right)}^2}} ,\quad 4 \leqslant t \leqslant 5} \end{array}} \right.\]

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so we have: \[\Large h\left( x \right) = \int_0^{2x - 1} {f\left( t \right)\;dt} \]
now, we have to distinguish these cases: \[0 \leqslant 2x - 1 < 4\]
and: \[4 \leqslant 2x - 1 \leqslant 6\]
first case: \[1 \leqslant x < \frac{5}{2}\]
we have: \[\large h\left( x \right) = \int_0^{2x - 1} {f\left( t \right)\;dt} = \int_0^{2x - 1} {\sqrt {4 - {{\left( {t - 2} \right)}^2}} \;dt} \]
second case: \[\frac{5}{2} \leqslant x \leqslant \frac{7}{2}\]
here we can write: \[\large h\left( x \right) = \int_0^{2x - 1} {f\left( t \right)\;dt} = - \int_0^{2x - 1} {\sqrt {1 - {{\left( {t - 5} \right)}^2}} \;dt} \]
both integral can be solved using a trigonometric substitution, namely, we have to substitute \[t - 2 = 2\sin \theta \] for first integral, and \[t - 5 = \sin \theta \] for the second one
integrals*
I'm computing the first integral...
please wait...
for first integral I got this expression: \[2\left\{ {\arcsin \left( {\frac{{2x - 3}}{2}} \right) + \frac{{2x - 3}}{2}\sqrt {1 - {{\left( {\frac{{2x - 3}}{2}} \right)}^2}} } \right\}\]
But we want the second integral to solve for h'(5/2), yes? Am I right about that?
yes! Nevertheless we have to determine the domain of both integrals
now the first integral can exist if and only if: \[ - 1 \leqslant \frac{{2x - 3}}{2} \leqslant 1\]
from which we get: \[\frac{1}{2} \leqslant x \leqslant \frac{5}{2}\]
so only between x=1/2 and x=2.5? So is that its domain?
now let's go to compute the second integral
The first integral is computed neglecting a constant term
for second integral I got this: \[\begin{gathered} h\left( x \right) = \int_5^{2x - 1} {f\left( t \right)\;dt} = - \int_5^{2x - 1} {\sqrt {1 - {{\left( {t - 5} \right)}^2}} \;dt} = \hfill \\ \hfill \\ = \frac{1}{2}\left\{ {\arcsin \left( {2x - 6} \right) + \left( {2x - 6} \right)\sqrt {1 - {{\left( {2x - 6} \right)}^2}} } \right\} \hfill \\ \hfill \\ \end{gathered} \]
your answer is right, the interval (0.5, 2.5) is part of the domain of h(x)
now the second integral exists if and only if: \[ - 1 \leqslant 2x - 6 \leqslant 1\]
But that's only for the part associated with the first integral, correct? The whole domain is 1/2<=x<=7/2
or equivalently: \[\frac{5}{2} \leqslant x \leqslant \frac{7}{2}\]
that's right, the domain of h(x) is given by the union of both sets, namely: \[\frac{1}{2} \leqslant x \leqslant \frac{7}{2}\]
now we have to determine h'(5/2)
using the fundamental theorem of integral calculus, we can write this:
we have to compute directly the first derivative of both expressions above
I'm computing, please wait...
Michele, what a great explanation. TSV, please be sure to rate when done, so Michele can get their share of OwlBucks
I got this: \[\Large h'\left( x \right) = \frac{4}{{\sqrt {1 - {{\left( {\frac{{2x - 3}}{2}} \right)}^2}} }},\quad \frac{1}{2} \leqslant x < \frac{5}{2}\quad \]
Thanks! Mrs. Preetha
now I compute h'(x) for the second expression:
@preetha I definitely will! I'm very grateful for the help on this problem, I'm doing some calc review and struggled with this concept! So this is very helpful for me!
please wait, I'm computing...
\[h'\left( x \right) = 2\sqrt {1 - {{\left( {2x - 6} \right)}^2}} ,\quad \frac{5}{2} \leqslant x \leqslant \frac{7}{2}\]
sorry the first equation for h'(x) is wrong, here is the right equation: \[h'\left( x \right) = 4\sqrt {1 - {{\left( {\frac{{2x - 3}}{2}} \right)}^2}} ,\quad \frac{1}{2} \leqslant x < \frac{5}{2}\quad \]
now, in order to find h'(5/2), we have to evaluate both expression above at x=5/2. So if we substitute x=5/2, we get: \[h'\left( {\frac{5}{2}} \right) = 4\sqrt {1 - {{\left( {\frac{{2 \times \frac{5}{2} - 3}}{2}} \right)}^2}} = 0\]
whereas the second expression, gives: \[h'\left( {\frac{5}{2}} \right) = 2\sqrt {1 - {{\left( {2 \times \frac{5}{2} - 6} \right)}^2}} = 0\]
so we can write: \[h'\left( {\frac{5}{2}} \right) = 0,\] in the ordinary sense of the first derivative
now let's go to the third part of your question
here we have to solve these equations: \[h'\left( x \right) = 4\sqrt {1 - {{\left( {\frac{{2x - 3}}{2}} \right)}^2}} = 0,\quad \frac{1}{2} \leqslant x < \frac{5}{2}\quad \]
and \[h'\left( x \right) = 2\sqrt {1 - {{\left( {2x - 6} \right)}^2}} = 0,\quad \frac{5}{2} \leqslant x \leqslant \frac{7}{2}\]
oops.. the second equation is: \[\begin{gathered} h'\left( x \right) = - 2\sqrt {1 - {{\left( {2x - 6} \right)}^2}} = 0,\quad \frac{5}{2} \leqslant x \leqslant \frac{7}{2} \hfill \\ \hfill \\ \end{gathered} \]
since we have: \[\begin{gathered} h\left( x \right) = \int_5^{2x - 1} {f\left( t \right)\;dt} = - \int_5^{2x - 1} {\sqrt {1 - {{\left( {t - 5} \right)}^2}} \;dt} = \hfill \\ \hfill \\ = - \frac{1}{2}\left\{ {\arcsin \left( {2x - 6} \right) + \left( {2x - 6} \right)\sqrt {1 - {{\left( {2x - 6} \right)}^2}} } \right\} \hfill \\ \end{gathered} \]
I'm pondering...
we have this result: the function: \[h'\left( x \right) = 4\sqrt {1 - {{\left( {\frac{{2x - 3}}{2}} \right)}^2}} \] is equal to zero at x=5/2 and x=1/2, furthermore it is always positive, so the function h(x) in that subintervals is always an increasing function
subinterval*
whereas, the function: \[h'\left( x \right) = - 2\sqrt {1 - {{\left( {2x - 6} \right)}^2}} \] is equal to zero at x=5/2 and at x=7/2. Furthermore it is always negative in that subinterval. So the function h(x) in that subinterval is a decreasing function
qualitatively, our function is represented by this drawing: |dw:1435090020580:dw|
namely x=5/2 is a local mximum for our function h(x)
Thank you so, so much! This has helped me a ton!!
:)
more precisely, the second integral, is: \[h\left( x \right) = \int_4^{2x - 1} {f\left( t \right)\;dt} = - \int_4^{2x - 1} {\sqrt {1 - {{\left( {t - 5} \right)}^2}} \;dt} = \] and it was computed neglecting a constant term, which doesn't affect the value of the first derivative

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