## anonymous one year ago 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

1. Michele_Laino

is your function like this: |dw:1435084959591:dw|

2. anonymous

Yes

3. Michele_Laino

we can write this: $\Large f\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {\sqrt {4 - {{\left( {t - 2} \right)}^2}} ,\quad 0 \leqslant t \leqslant 4} \\ { - \sqrt {1 - {{\left( {t - 5} \right)}^2}} ,\quad 4 \leqslant t \leqslant 5} \end{array}} \right.$

4. Michele_Laino

so we have: $\Large h\left( x \right) = \int_0^{2x - 1} {f\left( t \right)\;dt}$

5. Michele_Laino

now, we have to distinguish these cases: $0 \leqslant 2x - 1 < 4$

6. Michele_Laino

and: $4 \leqslant 2x - 1 \leqslant 6$

7. Michele_Laino

first case: $1 \leqslant x < \frac{5}{2}$

8. Michele_Laino

we have: $\large h\left( x \right) = \int_0^{2x - 1} {f\left( t \right)\;dt} = \int_0^{2x - 1} {\sqrt {4 - {{\left( {t - 2} \right)}^2}} \;dt}$

9. Michele_Laino

second case: $\frac{5}{2} \leqslant x \leqslant \frac{7}{2}$

10. Michele_Laino

here we can write: $\large h\left( x \right) = \int_0^{2x - 1} {f\left( t \right)\;dt} = - \int_0^{2x - 1} {\sqrt {1 - {{\left( {t - 5} \right)}^2}} \;dt}$

11. Michele_Laino

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

12. Michele_Laino

integrals*

13. Michele_Laino

I'm computing the first integral...

14. Michele_Laino

15. Michele_Laino

for first integral I got this expression: $2\left\{ {\arcsin \left( {\frac{{2x - 3}}{2}} \right) + \frac{{2x - 3}}{2}\sqrt {1 - {{\left( {\frac{{2x - 3}}{2}} \right)}^2}} } \right\}$

16. anonymous

But we want the second integral to solve for h'(5/2), yes? Am I right about that?

17. Michele_Laino

yes! Nevertheless we have to determine the domain of both integrals

18. Michele_Laino

now the first integral can exist if and only if: $- 1 \leqslant \frac{{2x - 3}}{2} \leqslant 1$

19. Michele_Laino

from which we get: $\frac{1}{2} \leqslant x \leqslant \frac{5}{2}$

20. anonymous

so only between x=1/2 and x=2.5? So is that its domain?

21. Michele_Laino

now let's go to compute the second integral

22. Michele_Laino

The first integral is computed neglecting a constant term

23. Michele_Laino

for second integral I got this: $\begin{gathered} h\left( x \right) = \int_5^{2x - 1} {f\left( t \right)\;dt} = - \int_5^{2x - 1} {\sqrt {1 - {{\left( {t - 5} \right)}^2}} \;dt} = \hfill \\ \hfill \\ = \frac{1}{2}\left\{ {\arcsin \left( {2x - 6} \right) + \left( {2x - 6} \right)\sqrt {1 - {{\left( {2x - 6} \right)}^2}} } \right\} \hfill \\ \hfill \\ \end{gathered}$

24. Michele_Laino

your answer is right, the interval (0.5, 2.5) is part of the domain of h(x)

25. Michele_Laino

now the second integral exists if and only if: $- 1 \leqslant 2x - 6 \leqslant 1$

26. anonymous

But that's only for the part associated with the first integral, correct? The whole domain is 1/2<=x<=7/2

27. Michele_Laino

or equivalently: $\frac{5}{2} \leqslant x \leqslant \frac{7}{2}$

28. Michele_Laino

that's right, the domain of h(x) is given by the union of both sets, namely: $\frac{1}{2} \leqslant x \leqslant \frac{7}{2}$

29. Michele_Laino

now we have to determine h'(5/2)

30. Michele_Laino

using the fundamental theorem of integral calculus, we can write this:

31. Michele_Laino

we have to compute directly the first derivative of both expressions above

32. Michele_Laino

I'm computing, please wait...

33. Preetha

Michele, what a great explanation. TSV, please be sure to rate when done, so Michele can get their share of OwlBucks

34. Michele_Laino

I got this: $\Large h'\left( x \right) = \frac{4}{{\sqrt {1 - {{\left( {\frac{{2x - 3}}{2}} \right)}^2}} }},\quad \frac{1}{2} \leqslant x < \frac{5}{2}\quad$

35. Michele_Laino

Thanks! Mrs. Preetha

36. Michele_Laino

now I compute h'(x) for the second expression:

37. anonymous

@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!

38. Michele_Laino

please wait, I'm computing...

39. Michele_Laino

$h'\left( x \right) = 2\sqrt {1 - {{\left( {2x - 6} \right)}^2}} ,\quad \frac{5}{2} \leqslant x \leqslant \frac{7}{2}$

40. Michele_Laino

sorry the first equation for h'(x) is wrong, here is the right equation: $h'\left( x \right) = 4\sqrt {1 - {{\left( {\frac{{2x - 3}}{2}} \right)}^2}} ,\quad \frac{1}{2} \leqslant x < \frac{5}{2}\quad$

41. Michele_Laino

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: $h'\left( {\frac{5}{2}} \right) = 4\sqrt {1 - {{\left( {\frac{{2 \times \frac{5}{2} - 3}}{2}} \right)}^2}} = 0$

42. Michele_Laino

whereas the second expression, gives: $h'\left( {\frac{5}{2}} \right) = 2\sqrt {1 - {{\left( {2 \times \frac{5}{2} - 6} \right)}^2}} = 0$

43. Michele_Laino

so we can write: $h'\left( {\frac{5}{2}} \right) = 0,$ in the ordinary sense of the first derivative

44. Michele_Laino

now let's go to the third part of your question

45. Michele_Laino

here we have to solve these equations: $h'\left( x \right) = 4\sqrt {1 - {{\left( {\frac{{2x - 3}}{2}} \right)}^2}} = 0,\quad \frac{1}{2} \leqslant x < \frac{5}{2}\quad$

46. Michele_Laino

and $h'\left( x \right) = 2\sqrt {1 - {{\left( {2x - 6} \right)}^2}} = 0,\quad \frac{5}{2} \leqslant x \leqslant \frac{7}{2}$

47. Michele_Laino

oops.. the second equation is: $\begin{gathered} h'\left( x \right) = - 2\sqrt {1 - {{\left( {2x - 6} \right)}^2}} = 0,\quad \frac{5}{2} \leqslant x \leqslant \frac{7}{2} \hfill \\ \hfill \\ \end{gathered}$

48. Michele_Laino

since we have: $\begin{gathered} h\left( x \right) = \int_5^{2x - 1} {f\left( t \right)\;dt} = - \int_5^{2x - 1} {\sqrt {1 - {{\left( {t - 5} \right)}^2}} \;dt} = \hfill \\ \hfill \\ = - \frac{1}{2}\left\{ {\arcsin \left( {2x - 6} \right) + \left( {2x - 6} \right)\sqrt {1 - {{\left( {2x - 6} \right)}^2}} } \right\} \hfill \\ \end{gathered}$

49. Michele_Laino

I'm pondering...

50. Michele_Laino

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

51. Michele_Laino

subinterval*

52. Michele_Laino

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

53. Michele_Laino

qualitatively, our function is represented by this drawing: |dw:1435090020580:dw|

54. Michele_Laino

namely x=5/2 is a local mximum for our function h(x)

55. anonymous

Thank you so, so much! This has helped me a ton!!

56. Michele_Laino

:)

57. Michele_Laino

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