Calculus Help

- anonymous

Calculus Help

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

Write \[\lim_{n \rightarrow \infty}\sum_{k=1}^{n}(2+k*(5/n))^3 *5/n\] as a definite Integral

- anonymous

Find \[dx/dy \int\limits_{2}^{x^3}\ln(x^2)dx\]

- anonymous

Find the area bounded by the curves y^2= 2x+6 and x=y+1. Work must be an integral with one variable

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- 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

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

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

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

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

for iluvhomewurk's did they just confuse 3 with 7?

- 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

the intersection point are:
\[\begin{gathered}
P = \left( { - 1, - 2} \right) \hfill \\
Q = \left( {5,4} \right) \hfill \\
\end{gathered} \]

- 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

are you still talking about question 3?

- Michele_Laino

yes!

- 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

|dw:1433964888030:dw|

- anonymous

so the answer is 28?

- Michele_Laino

yes!

- Michele_Laino

oops.. it is 18

- Michele_Laino

sorry!

- 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

its okay
Is the work that iluvhomework did correct?

- Michele_Laino

please wait I'm checking...

- anonymous

okay thank you. I think it is just its 7 instead of 3

- anonymous

and i got 0 not 18

- Michele_Laino

area=0?

- anonymous

the step is to plug in 4 into equation then subtract the equation with -2 pulgged into it correct?

- Michele_Laino

yes! correct!

- anonymous

When i did that i got 0

- Michele_Laino

I'm checking....

- 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

\[\large \int_{ - 2}^4 {dy\left( {y + 1 - \frac{{{y^2} - 6}}{2}} \right)} = \left. {\frac{{{y^2}}}{2} - \frac{{{y^3}}}{6} + 4y} \right|_{ - 2}^4 = 18\]

- Michele_Laino

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

as you can see my new reference system has the Y-axis along the axis of rotation

- Michele_Laino

|dw:1433966377727:dw|

- 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

what problem is this for? the one that ilovehomework started?

- Michele_Laino

here is our situation:
|dw:1433966625656:dw|

- Michele_Laino

yes!

- anonymous

Im confused on what work i am supposed to show

- anonymous

is ilovehomeworks work not correct?

- Michele_Laino

I give my solution, after that you can choose the solution which you prefer

- Michele_Laino

I got a different result

- anonymous

okay then please outline step by step on how you did the problem. Tnak you

- anonymous

okay then please outline step by step on how you did the problem. Tnak you

- Michele_Laino

ok! Now our volume is given integrating the area of annulus like below, along the Y-axis:
|dw:1433967160201:dw|

- anonymous

okay how do you show that mathematically.

- 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

so thats my first step to write down?

- Michele_Laino

your first step are the equations of traslation, I think

- anonymous

okay

- 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

and the final result is:
\[\Large V = 8\pi \]

- anonymous

- 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

|dw:1433968218028:dw|

- anonymous

okay

- 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

so we use both equations?

- Michele_Laino

yes!

- 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

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

so:
\[\Large \tau = \frac{{{v_0}}}{g}\]

- anonymous

so t=40/-32?

- Michele_Laino

no, it is \tau = 40/32=...

- Michele_Laino

g is the magnitude of gravity: g=9.81 m/sec^2=32 feet/sec^2

- anonymous

wont it be -32?

- 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

where h=500 feet

- 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

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

\[\large \begin{gathered}
v = \sqrt {2gz\left( \tau \right)} = \sqrt {2gh + v_0^2} = \sqrt {2 \times 32 \times 500 + {{40}^2}} = \hfill \\
= 183.3\;{\text{feet/sec}} \hfill \\
\end{gathered} \]

- anonymous

and thats the final answer?

- Michele_Laino

yes!

- anonymous

Thank you so much. Could you help me with about 3 more?

- Michele_Laino

ok!

- 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

please wait I'm pondering...

- anonymous

okay thank you

- 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

|dw:1433970078818:dw|

- anonymous

ok i see the intersection points

- Michele_Laino

so, it is suffice to compute the bounded area which lies into the first quadrant

- anonymous

yes I agree with that

- Michele_Laino

|dw:1433970288453:dw|

- 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

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

simplidy it to 7/32?

- Michele_Laino

|dw:1433970540095:dw|
the area of bounded region is:
\[\Large {A_B} = \frac{1}{2} - \frac{{14}}{{64}} = \frac{7}{{16}}\]

- anonymous

oh okay

- Michele_Laino

so the requested area is:
\[\Large 2{A_B} = \frac{7}{8}\]

- anonymous

who do you format it like that?

- anonymous

why

- Michele_Laino

since our drawing is symmetric with respect to the y-axis:
|dw:1433970857126:dw|

- anonymous

wouldnt one of the areas be negative because it is below the x axis?

- anonymous

so it would be 0?

- 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

okay so the fonal answer would be 7/8?

- Michele_Laino

yes!

- anonymous

okay heres the next one

- Michele_Laino

ok!

- anonymous

I have completed it i just wanna make sure i did it correct.

- 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

t 0 1 3 5 6
R(t) 4 2 3 2 1

- 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

I think i did it correctly

- Michele_Laino

please wait, I'm checking your computation...

- anonymous

- anonymous

sorry dont know why it sent again

- 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

okat thank you for al this help just two more!

- Michele_Laino

ok!

- anonymous

Show that the function \[F(x)= \int\limits_{x}^{3x}1/t dt\] is constant on the interval(0, infty)

- 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

how did you get ln(3x) and ln(x) from the integral?

- Michele_Laino

yes! since we can write:
\[\Large \int {\frac{{dt}}{t}} = \ln t + k\]

- anonymous

oh okay!

- anonymous

and since it comes out to be a number that means it is constant?

- Michele_Laino

correct!

- anonymous

perfect! last question

- Michele_Laino

ok!

- anonymous

write\[\lim_{n \rightarrow \infty} \sum_{k=1}^{n} (2+k*5/n)^3*5/n\] as a definite integral

- Michele_Laino

I'm pondering...

- anonymous

Thank you

- Michele_Laino

what is k

- anonymous

constant

- Michele_Laino

is it a parameter?

- Michele_Laino

ok!

- anonymous

I think i got it

- anonymous

Except 2+x instead of 2+5x

##### 1 Attachment

- anonymous

does that look correct?

- 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

Thank you for all your help i am finished

- 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

and finally, I get:
\[\Large \begin{gathered}
\int_0^n {dk} {\left( {2 + \frac{{5k}}{n}} \right)^3}\frac{5}{n} = \left. {{{\left( {2 + \frac{{5k}}{n}} \right)}^4}\frac{1}{4}} \right|_0^n = \hfill \\
\hfill \\
= \frac{{{7^4} - {2^4}}}{4} \hfill \\
\end{gathered} \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.