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anonymous
 one year ago
Calculus Help
anonymous
 one year ago
Calculus Help

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Write \[\lim_{n \rightarrow \infty}\sum_{k=1}^{n}(2+k*(5/n))^3 *5/n\] as a definite Integral

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Find \[dx/dy \int\limits_{2}^{x^3}\ln(x^2)dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Find the area bounded by the curves y^2= 2x+6 and x=y+1. Work must be an integral with one variable

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Write 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
 one year ago
Best ResponseYou've already chosen the best response.00.5x² = x 0.5x²  x = 0 x (0.5x  1) = 0 x = 0, 2 Those are the xcoordinates 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 3x (because x is the distance from the yaxis to the outer edge of the shell, leaving you with 3x 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∫ (3x)(x  0.5x²) dx, from x = 0 to 2 = 2π∫ (3xx²  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
 one year ago
Best ResponseYou've already chosen the best response.0A 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
 one year ago
Best ResponseYou've already chosen the best response.10question #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
 one year ago
Best ResponseYou've already chosen the best response.10so 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
 one year ago
Best ResponseYou've already chosen the best response.0for iluvhomewurk's did they just confuse 3 with 7?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10Question #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
 one year ago
Best ResponseYou've already chosen the best response.10the intersection point are: \[\begin{gathered} P = \left( {  1,  2} \right) \hfill \\ Q = \left( {5,4} \right) \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10so the requested area, is given by the subsequent integral: \[\Large \int_{  2}^4 {dy\left( {y + 1  \frac{{{y^2}  6}}{2}} \right)} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0are you still talking about question 3?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10here 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
 one year ago
Best ResponseYou've already chosen the best response.10dw:1433964888030:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the answer is 28?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10oops.. it is 18

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10\[\Large \int_{  2}^4 {dy\left( {y + 1  \frac{{{y^2}  6}}{2}} \right)} = \left. {y  \frac{{{y^2}}}{2} + 4} \right_{  2}^4 = 18\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its okay Is the work that iluvhomework did correct?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10please wait I'm checking...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay thank you. I think it is just its 7 instead of 3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the step is to plug in 4 into equation then subtract the equation with 2 pulgged into it correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0When i did that i got 0

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10I'm checking....

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10you 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
 one year ago
Best ResponseYou've already chosen the best response.10\[\large \int_{  2}^4 {dy\left( {y + 1  \frac{{{y^2}  6}}{2}} \right)} = \left. {\frac{{{y^2}}}{2}  \frac{{{y^3}}}{6} + 4y} \right_{  2}^4 = 18\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10question #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
 one year ago
Best ResponseYou've already chosen the best response.10as you can see my new reference system has the Yaxis along the axis of rotation

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10dw:1433966377727:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10so 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
 one year ago
Best ResponseYou've already chosen the best response.0what problem is this for? the one that ilovehomework started?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10here is our situation: dw:1433966625656:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im confused on what work i am supposed to show

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is ilovehomeworks work not correct?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10I give my solution, after that you can choose the solution which you prefer

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10I got a different result

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay then please outline step by step on how you did the problem. Tnak you

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay then please outline step by step on how you did the problem. Tnak you

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10ok! Now our volume is given integrating the area of annulus like below, along the Yaxis: dw:1433967160201:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay how do you show that mathematically.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10here 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
 one year ago
Best ResponseYou've already chosen the best response.0so thats my first step to write down?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10your first step are the equations of traslation, I think

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10now, 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
 one year ago
Best ResponseYou've already chosen the best response.10and the final result is: \[\Large V = 8\pi \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A 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
 one year ago
Best ResponseYou've already chosen the best response.10here 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
 one year ago
Best ResponseYou've already chosen the best response.10dw:1433968218028:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10furthermore, we have to consider this one: \[\Large {\mathbf{v}}\left( t \right) = {{\mathbf{v}}_{\mathbf{0}}} + {\mathbf{g}}t\] dw:1433968355520:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we use both equations?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10now I rewrite both equations using zcomponents: \[\Large \left\{ \begin{gathered} z\left( t \right) = {z_0} + {v_0}t  \frac{1}{2}g{t^2} \hfill \\ {v_z}\left( t \right) = {v_0}  gt \hfill \\ \end{gathered} \right.\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10when 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
 one year ago
Best ResponseYou've already chosen the best response.10so: \[\Large \tau = \frac{{{v_0}}}{g}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10no, it is \tau = 40/32=...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10g is the magnitude of gravity: g=9.81 m/sec^2=32 feet/sec^2

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10no, 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
 one year ago
Best ResponseYou've already chosen the best response.10where h=500 feet

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10next, 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
 one year ago
Best ResponseYou've already chosen the best response.10substituting 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
 one year ago
Best ResponseYou've already chosen the best response.10\[\large \begin{gathered} v = \sqrt {2gz\left( \tau \right)} = \sqrt {2gh + v_0^2} = \sqrt {2 \times 32 \times 500 + {{40}^2}} = \hfill \\ = 183.3\;{\text{feet/sec}} \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and thats the final answer?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much. Could you help me with about 3 more?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Use the midpoint rule with n=4 to approximate the area of the region bounded by y=x^3 and y=x

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10please wait I'm pondering...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10the 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
 one year ago
Best ResponseYou've already chosen the best response.10dw:1433970078818:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok i see the intersection points

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10so, it is suffice to compute the bounded area which lies into the first quadrant

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes I agree with that

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10dw:1433970288453:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10the 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
 one year ago
Best ResponseYou've already chosen the best response.10so 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
 one year ago
Best ResponseYou've already chosen the best response.0simplidy it to 7/32?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10dw:1433970540095:dw the area of bounded region is: \[\Large {A_B} = \frac{1}{2}  \frac{{14}}{{64}} = \frac{7}{{16}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10so the requested area is: \[\Large 2{A_B} = \frac{7}{8}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0who do you format it like that?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10since our drawing is symmetric with respect to the yaxis: dw:1433970857126:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wouldnt one of the areas be negative because it is below the x axis?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10no, since an area is the measure of the subset of the x,yplane, and the measure of a set, as it is defined by mathematicians, is always positive

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so the fonal answer would be 7/8?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay heres the next one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have completed it i just wanna make sure i did it correct.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Trash 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
 one year ago
Best ResponseYou've already chosen the best response.0t 0 1 3 5 6 R(t) 4 2 3 2 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Lsum= (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
 one year ago
Best ResponseYou've already chosen the best response.0I think i did it correctly

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10please wait, I'm checking your computation...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Lsum= (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
 one year ago
Best ResponseYou've already chosen the best response.0sorry dont know why it sent again

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10Lsum 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
 one year ago
Best ResponseYou've already chosen the best response.0okat thank you for al this help just two more!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Show that the function \[F(x)= \int\limits_{x}^{3x}1/t dt\] is constant on the interval(0, infty)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10it 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
 one year ago
Best ResponseYou've already chosen the best response.0how did you get ln(3x) and ln(x) from the integral?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10yes! since we can write: \[\Large \int {\frac{{dt}}{t}} = \ln t + k\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and since it comes out to be a number that means it is constant?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0perfect! last question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0write\[\lim_{n \rightarrow \infty} \sum_{k=1}^{n} (2+k*5/n)^3*5/n\] as a definite integral

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10I'm pondering...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10is it a parameter?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Except 2+x instead of 2+5x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0does that look correct?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10here 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
 one year ago
Best ResponseYou've already chosen the best response.0Thank you for all your help i am finished

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.10which 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
 one year ago
Best ResponseYou've already chosen the best response.10and 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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