anonymous
  • anonymous
Calculus Help
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
Write \[\lim_{n \rightarrow \infty}\sum_{k=1}^{n}(2+k*(5/n))^3 *5/n\] as a definite Integral
anonymous
  • anonymous
Find \[dx/dy \int\limits_{2}^{x^3}\ln(x^2)dx\]
anonymous
  • anonymous
Find the area bounded by the curves y^2= 2x+6 and x=y+1. Work must be an integral with one variable

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anonymous
  • anonymous
Write the integral in one variable to find the volume of the solid obtained by rotating the first quadrant region bounded by y=0.5x^2 and y=x about the line x=7.
anonymous
  • anonymous
0.5x² = x 0.5x² - x = 0 x (0.5x - 1) = 0 x = 0, 2 Those are the x-coordinates of the two points of intersection of y = 00.5x² and y=x. The full points are (0,0) and (2,2). Revolving the region around a vertical axis like x = 3 is better suited to the cylindrical shell method so that you don't have to solve for y. You integrate the lateral surface area of a cylindrical shell having its height vertically within the bounded region. The radius is 3-x (because x is the distance from the y-axis to the outer edge of the shell, leaving you with 3-x as the distance from the outer edge of the shell to x=3). The height of the cylinder is upper curve minus lower curve = x - 0.5x². Therefore, volume = 2π∫rh dx = 2π∫ (3-x)(x - 0.5x²) dx, from x = 0 to 2 = 2π∫ (3x-x² - 1.5x² + 0.5x³) dx, from x = 0 to 2 = 2π∫ (3x - 2.5x² + 0.5x³) dx, from x = 0 to 2 = 2π [3x²/2 - 2.5x³/3 + 0.5x⁴/4], from x = 0 to 2 = 2π [6 - 20/3 + 2] = 2π [4/3] = 8π/3
anonymous
  • anonymous
A ball is thrown upward with a speed of 40 feet per second from the edge of a cliff 500 feet above the ground. What is the speed of the ball when it hits the ground? Use acceleration due to gravity as -32 feet per second squared.
Michele_Laino
  • Michele_Laino
question #2: the integral is, integrating by parts, is: \[\Large \begin{gathered} \int {dx\ln \left( {{x^2}} \right)} = x\ln \left( {{x^2}} \right) - \int {x\frac{{dx}}{{{x^2}}}} 2x = \hfill \\ \hfill \\ = x\ln \left( {{x^2}} \right) - 2x \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
so the definite integral is: \[\Large \left. {x\ln \left( {{x^2}} \right) - 2x} \right|_2^{{x^3}} = {x^3}\ln \left( {{x^6}} \right) - 2{x^3} - 2\ln 4 + 4\]
anonymous
  • anonymous
for iluvhomewurk's did they just confuse 3 with 7?
Michele_Laino
  • Michele_Laino
Question #3 we can rewrite your equations as follows: \[\Large \left\{ \begin{gathered} x = \frac{{{y^2} - 6}}{2} \hfill \\ x = y + 1 \hfill \\ \end{gathered} \right.\]
Michele_Laino
  • Michele_Laino
the intersection point are: \[\begin{gathered} P = \left( { - 1, - 2} \right) \hfill \\ Q = \left( {5,4} \right) \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
so the requested area, is given by the subsequent integral: \[\Large \int_{ - 2}^4 {dy\left( {y + 1 - \frac{{{y^2} - 6}}{2}} \right)} \]
anonymous
  • anonymous
are you still talking about question 3?
Michele_Laino
  • Michele_Laino
yes!
Michele_Laino
  • Michele_Laino
here is the next step: \[\Large \int_{ - 2}^4 {dy\left( {y + 1 - \frac{{{y^2} - 6}}{2}} \right)} = \left. {y - \frac{{{y^2}}}{2} + 4} \right|_{ - 2}^4 = 28\]
Michele_Laino
  • Michele_Laino
|dw:1433964888030:dw|
anonymous
  • anonymous
so the answer is 28?
Michele_Laino
  • Michele_Laino
yes!
Michele_Laino
  • Michele_Laino
oops.. it is 18
Michele_Laino
  • Michele_Laino
sorry!
Michele_Laino
  • Michele_Laino
\[\Large \int_{ - 2}^4 {dy\left( {y + 1 - \frac{{{y^2} - 6}}{2}} \right)} = \left. {y - \frac{{{y^2}}}{2} + 4} \right|_{ - 2}^4 = 18\]
anonymous
  • anonymous
its okay Is the work that iluvhomework did correct?
Michele_Laino
  • Michele_Laino
please wait I'm checking...
anonymous
  • anonymous
okay thank you. I think it is just its 7 instead of 3
anonymous
  • anonymous
and i got 0 not 18
Michele_Laino
  • Michele_Laino
area=0?
anonymous
  • anonymous
the step is to plug in 4 into equation then subtract the equation with -2 pulgged into it correct?
Michele_Laino
  • Michele_Laino
yes! correct!
anonymous
  • anonymous
When i did that i got 0
Michele_Laino
  • Michele_Laino
I'm checking....
Michele_Laino
  • Michele_Laino
you are right, I gave you the integrand function, please here is the integral: \[\Large \int_{ - 2}^4 {dy\left( {y + 1 - \frac{{{y^2} - 6}}{2}} \right)} = \left. {\frac{{{y^2}}}{2} - \frac{{{y^3}}}{6} + 4y} \right|_{ - 2}^4 = 18\]
Michele_Laino
  • Michele_Laino
\[\large \int_{ - 2}^4 {dy\left( {y + 1 - \frac{{{y^2} - 6}}{2}} \right)} = \left. {\frac{{{y^2}}}{2} - \frac{{{y^3}}}{6} + 4y} \right|_{ - 2}^4 = 18\]
Michele_Laino
  • Michele_Laino
question #3 I make this traslation: \[\Large \left\{ \begin{gathered} x = X + 7 \hfill \\ y = Y \hfill \\ \end{gathered} \right.\] where X, and Y are the new coordinates
Michele_Laino
  • Michele_Laino
as you can see my new reference system has the Y-axis along the axis of rotation
Michele_Laino
  • Michele_Laino
|dw:1433966377727:dw|
Michele_Laino
  • Michele_Laino
so the equations of your lines, becomes: \[\Large \begin{gathered} y = \frac{{{x^2}}}{2} \to Y = \frac{{{{\left( {X + 7} \right)}^2}}}{2} \hfill \\ \hfill \\ y = x \to Y = X + 7 \hfill \\ \end{gathered} \]
anonymous
  • anonymous
what problem is this for? the one that ilovehomework started?
Michele_Laino
  • Michele_Laino
here is our situation: |dw:1433966625656:dw|
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
Im confused on what work i am supposed to show
anonymous
  • anonymous
is ilovehomeworks work not correct?
Michele_Laino
  • Michele_Laino
I give my solution, after that you can choose the solution which you prefer
Michele_Laino
  • Michele_Laino
I got a different result
anonymous
  • anonymous
okay then please outline step by step on how you did the problem. Tnak you
anonymous
  • anonymous
okay then please outline step by step on how you did the problem. Tnak you
Michele_Laino
  • Michele_Laino
ok! Now our volume is given integrating the area of annulus like below, along the Y-axis: |dw:1433967160201:dw|
anonymous
  • anonymous
okay how do you show that mathematically.
Michele_Laino
  • Michele_Laino
here is the integral which gives the requested volume: \[\Large \int_0^2 {dY\left\{ {{{\left( { - 7 + Y} \right)}^2} - {{\left( { - 7 + \sqrt {2Y} } \right)}^2}} \right\}} \pi \]
anonymous
  • anonymous
so thats my first step to write down?
Michele_Laino
  • Michele_Laino
your first step are the equations of traslation, I think
anonymous
  • anonymous
okay
Michele_Laino
  • Michele_Laino
now, developing that integral, we get: \[\Large \begin{gathered} \int_0^2 {dY\left\{ {{{\left( { - 7 + Y} \right)}^2} - {{\left( { - 7 + \sqrt {2Y} } \right)}^2}} \right\}} \pi = \hfill \\ \hfill \\ = \pi \left( {\left. {\frac{{{Y^3}}}{3} - 16\frac{{{Y^2}}}{2} + 14\sqrt 2 \times \frac{{2{Y^{3/2}}}}{3}} \right|_0^2} \right) \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
and the final result is: \[\Large V = 8\pi \]
anonymous
  • anonymous
A ball is thrown upward with a speed of 40 feet per second from the edge of a cliff 500 feet above the ground. What is the speed of the ball when it hits the ground? Use acceleration due to gravity as -32 feet per second squared.
Michele_Laino
  • Michele_Laino
here we have to use the subsequent vector equation: \[\Large {\mathbf{OP}}\left( t \right) = {\mathbf{O}}{{\mathbf{P}}_{\mathbf{0}}} + {{\mathbf{v}}_{\mathbf{0}}}t + \frac{1}{2}{\mathbf{g}}{t^2}\]
Michele_Laino
  • Michele_Laino
|dw:1433968218028:dw|
anonymous
  • anonymous
okay
Michele_Laino
  • Michele_Laino
furthermore, we have to consider this one: \[\Large {\mathbf{v}}\left( t \right) = {{\mathbf{v}}_{\mathbf{0}}} + {\mathbf{g}}t\] |dw:1433968355520:dw|
anonymous
  • anonymous
so we use both equations?
Michele_Laino
  • Michele_Laino
yes!
Michele_Laino
  • Michele_Laino
now I rewrite both equations using z-components: \[\Large \left\{ \begin{gathered} z\left( t \right) = {z_0} + {v_0}t - \frac{1}{2}g{t^2} \hfill \\ {v_z}\left( t \right) = {v_0} - gt \hfill \\ \end{gathered} \right.\]
Michele_Laino
  • Michele_Laino
when the ball reaches its maximum height, then its velocity is zero, so we can write: \[\Large 0 = {v_0} - g\tau \] where \tau is the time at which our ball reaches its maximum height
Michele_Laino
  • Michele_Laino
so: \[\Large \tau = \frac{{{v_0}}}{g}\]
anonymous
  • anonymous
so t=40/-32?
Michele_Laino
  • Michele_Laino
no, it is \tau = 40/32=...
Michele_Laino
  • Michele_Laino
g is the magnitude of gravity: g=9.81 m/sec^2=32 feet/sec^2
anonymous
  • anonymous
wont it be -32?
Michele_Laino
  • Michele_Laino
no, since I wrote my scalar equations using the magnitudes or length of the respective vectors the maximum height is then (substituting \tau into the first scalar equation): \[\Large \begin{gathered} z\left( \tau \right) = {z_0} + {v_0}\tau - \frac{1}{2}g{\tau ^2} = h + \frac{{v_0^2}}{g} - \frac{1}{2}\frac{{v_0^2}}{g} = \hfill \\ \hfill \\ = h + \frac{{v_0^2}}{{2g}} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
where h=500 feet
Michele_Laino
  • Michele_Laino
next, in order to find the requested velocity, we can apply the formula of Torricelli, like below: \[\Large v = \sqrt {2gz\left( \tau \right)} = \sqrt {2gh + v_0^2} \]
Michele_Laino
  • Michele_Laino
substituting your data we find: \[\Large \begin{gathered} v = \sqrt {2gz\left( \tau \right)} = \sqrt {2gh + v_0^2} = \sqrt {2 \times 32 \times 500 + {{40}^2}} = \hfill \\ = 183.3\;{\text{feet/sec}} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
\[\large \begin{gathered} v = \sqrt {2gz\left( \tau \right)} = \sqrt {2gh + v_0^2} = \sqrt {2 \times 32 \times 500 + {{40}^2}} = \hfill \\ = 183.3\;{\text{feet/sec}} \hfill \\ \end{gathered} \]
anonymous
  • anonymous
and thats the final answer?
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
Thank you so much. Could you help me with about 3 more?
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
Use the midpoint rule with n=4 to approximate the area of the region bounded by y=x^3 and y=x
Michele_Laino
  • Michele_Laino
please wait I'm pondering...
anonymous
  • anonymous
okay thank you
Michele_Laino
  • Michele_Laino
the intersection points between those equations are: \[\begin{gathered} P = \left( { - 1, - 1} \right) \hfill \\ Q = \left( {1,1} \right) \hfill \\ O = \left( {0,0} \right) \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
|dw:1433970078818:dw|
anonymous
  • anonymous
ok i see the intersection points
Michele_Laino
  • Michele_Laino
so, it is suffice to compute the bounded area which lies into the first quadrant
anonymous
  • anonymous
yes I agree with that
Michele_Laino
  • Michele_Laino
|dw:1433970288453:dw|
Michele_Laino
  • Michele_Laino
the area under the cubic parabola, is given by the subsequent formula: \[\Large A = \frac{{\left( {1 - 0} \right)}}{2}\left( {f\left( {1/4} \right) + f\left( {3/4} \right)} \right)\] where f(x) = x^3
Michele_Laino
  • Michele_Laino
so we get: \[\Large \begin{gathered} A = \frac{{\left( {1 - 0} \right)}}{2}\left( {f\left( {1/4} \right) + f\left( {3/4} \right)} \right) = \hfill \\ \hfill \\ = \frac{1}{2}\left\{ {\frac{1}{{64}} + \frac{{27}}{{64}}} \right\} = \frac{{14}}{{64}} \hfill \\ \end{gathered} \]
anonymous
  • anonymous
simplidy it to 7/32?
Michele_Laino
  • Michele_Laino
|dw:1433970540095:dw| the area of bounded region is: \[\Large {A_B} = \frac{1}{2} - \frac{{14}}{{64}} = \frac{7}{{16}}\]
anonymous
  • anonymous
oh okay
Michele_Laino
  • Michele_Laino
so the requested area is: \[\Large 2{A_B} = \frac{7}{8}\]
anonymous
  • anonymous
who do you format it like that?
anonymous
  • anonymous
why
Michele_Laino
  • Michele_Laino
since our drawing is symmetric with respect to the y-axis: |dw:1433970857126:dw|
anonymous
  • anonymous
wouldnt one of the areas be negative because it is below the x axis?
anonymous
  • anonymous
so it would be 0?
Michele_Laino
  • Michele_Laino
no, since an area is the measure of the subset of the x,y-plane, and the measure of a set, as it is defined by mathematicians, is always positive
anonymous
  • anonymous
okay so the fonal answer would be 7/8?
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
okay heres the next one
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
I have completed it i just wanna make sure i did it correct.
anonymous
  • anonymous
Trash is delivered to a dumpsite between the hours of midnight(t=0) and 6am (t=6). Selected Values of the rates of deleivery are shown in the table below with t measured in hours and r(t) measured in tons per hours. Use 4 trapezoids to estimate total amount of trash. Round to nearest ton
anonymous
  • anonymous
t 0 1 3 5 6 R(t) 4 2 3 2 1
anonymous
  • anonymous
Lsum= (1)(4)+(2)(2)+(2)(3)+(1)(2)=4+4+6+2= 16 Rsum= (1)(2)+(2)(3)+(2)(2)+(1)(1)= 2+6+4+1= 13 Tsum= ½(Rsum+Lsum)= ½(16+13)=1/2(29)= 14.5 15 Tons
anonymous
  • anonymous
I think i did it correctly
Michele_Laino
  • Michele_Laino
please wait, I'm checking your computation...
anonymous
  • anonymous
Lsum= (1)(4)+(2)(2)+(2)(3)+(1)(2)=4+4+6+2= 16 Rsum= (1)(2)+(2)(3)+(2)(2)+(1)(1)= 2+6+4+1= 13 Tsum= ½(Rsum+Lsum)= ½(16+13)=1/2(29)= 14.5 15 Tons
anonymous
  • anonymous
sorry dont know why it sent again
Michele_Laino
  • Michele_Laino
Lsum is right for me whereas I write Rsum like this: Rsum= (2)(2)+(2)(3)+(2)(2)+(1)(1)= 4+6+4+1= 15
anonymous
  • anonymous
okat thank you for al this help just two more!
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
Show that the function \[F(x)= \int\limits_{x}^{3x}1/t dt\] is constant on the interval(0, infty)
Michele_Laino
  • Michele_Laino
it is simple, since we have: \[\Large \int_x^{3x} {\frac{{dt}}{t}} = \ln \left( {3x} \right) - \ln \left( x \right) = \ln \left( {\frac{{3x}}{x}} \right) = \ln 3\]
anonymous
  • anonymous
how did you get ln(3x) and ln(x) from the integral?
Michele_Laino
  • Michele_Laino
yes! since we can write: \[\Large \int {\frac{{dt}}{t}} = \ln t + k\]
anonymous
  • anonymous
oh okay!
anonymous
  • anonymous
and since it comes out to be a number that means it is constant?
Michele_Laino
  • Michele_Laino
correct!
anonymous
  • anonymous
perfect! last question
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
write\[\lim_{n \rightarrow \infty} \sum_{k=1}^{n} (2+k*5/n)^3*5/n\] as a definite integral
Michele_Laino
  • Michele_Laino
I'm pondering...
anonymous
  • anonymous
Thank you
Michele_Laino
  • Michele_Laino
what is k
anonymous
  • anonymous
constant
Michele_Laino
  • Michele_Laino
is it a parameter?
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
I think i got it
anonymous
  • anonymous
Except 2+x instead of 2+5x
anonymous
  • anonymous
does that look correct?
Michele_Laino
  • Michele_Laino
here is my answer: I write the sum as integral, like below: \[\Large \int_0^n {dk} {\left( {2 + \frac{{5k}}{n}} \right)^3}\frac{5}{n}\]
anonymous
  • anonymous
Thank you for all your help i am finished
Michele_Laino
  • Michele_Laino
which is equal to: \[\Large \int_0^n {dk} {\left( {2 + \frac{{5k}}{n}} \right)^3}\frac{5}{n} = \left. {{{\left( {2 + \frac{{5k}}{n}} \right)}^4}\frac{1}{4}} \right|_0^n\]
Michele_Laino
  • Michele_Laino
and finally, I get: \[\Large \begin{gathered} \int_0^n {dk} {\left( {2 + \frac{{5k}}{n}} \right)^3}\frac{5}{n} = \left. {{{\left( {2 + \frac{{5k}}{n}} \right)}^4}\frac{1}{4}} \right|_0^n = \hfill \\ \hfill \\ = \frac{{{7^4} - {2^4}}}{4} \hfill \\ \end{gathered} \]

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