A few Integral problems

**DJ Hobo** · September 3rd 2009, 12:01 AM

Attached is a sheet with a few problems that I either can't start, or can't finish.

In question 10, I'm having trouble with parts c and d (although, if/when I get the answer to c, I think I should be able to get d). It looks a bit blurry, but the number above the integrand sign is 4 and the number below is 0, in case there's any confusion there.

In question 11, I think my main problem is I don't know the steps, or the approach, I need to take. The only thing I've done so far for this particular question is find $\int_{0}^{4}{\left( 4x\; -\; x^{2} \right)}dx$ (and I don't even know if that particular step is relevant or helpful).

As for question 12; I don't even understand the information given. Could someone please explain it in more simple terms, please?

Any help here is, of course, greatly appreciated.

**DeMath** · September 3rd 2009, 02:06 AM

Originally Posted by DJ Hobo

Attached is a sheet with a few problems that I either can't start, or can't finish.

In question 10, I'm having trouble with parts c and d (although, if/when I get the answer to c, I think I should be able to get d). It looks a bit blurry, but the number above the integrand sign is 4 and the number below is 0, in case there's any confusion there.

In question 11, I think my main problem is I don't know the steps, or the approach, I need to take. The only thing I've done so far for this particular question is find $\int_{0}^{4}{\left( 4x\; -\; x^{2} \right)}dx$ (and I don't even know if that particular step is relevant or helpful).

As for question 12; I don't even understand the information given. Could someone please explain it in more simple terms, please?

Any help here is, of course, greatly appreciated.

For the question 11

You have $\, {y_1} = ax, \, {y_2} = 4x - {x^2}.$

Find intersection

$ax = 4x - {x^2} \Leftrightarrow {x^2} - \left( {4 - a} \right)x = 0 \Leftrightarrow \left[ \begin{gathered} {x_1} = 0, \hfill \\ {x_2} = 4 - a. \hfill \\ \end{gathered} \right.$

Consequently, you got that $k = 4 - a \Leftrightarrow a = 4 - k$

By the condition of your task you have

$\int\limits_0^k {\left( {{y_2} - {y_1}} \right)dx} = \frac{1} {2}\int\limits_0^4 {{y_2}\,dx} ,$

$\int\limits_0^k {\left( {4x - {x^2} - ax} \right)dx} = \int\limits_0^k {\left( {4x - {x^2} - \left( {4 - k} \right)x} \right)dx} = \frac{1} {2}\int\limits_0^4 {\left( {4x - {x^2}} \right)dx} .$

$\int\limits_0^k {\left( {4x - {x^2} - \left( {4 - k} \right)x} \right)dx} = \int\limits_0^k {\left( {kx - {x^2}} \right)} dx = \left. {\left( {\frac{{k{x^2}}} {2} - \frac{{{x^3}}} {3}} \right)} \right|_0^k = \frac{{{k^3}}} {2} - \frac{{{k^3}}} {3} = \frac{{{k^3}}}{6}.$

$\int\limits_0^4 {\left( {4x - {x^2}} \right)dx} = \left. {\left( {2{x^2} - \frac{{{x^3}}} {3}} \right)} \right|_0^4 = 32 - \frac{{64}} {3} = \frac{{32}} {3}.$

Finally you have

$\frac{{{k^3}}}{6} = \frac{1}{2} \cdot \frac{{32}}{3} \Leftrightarrow {k^3} = 32 \Rightarrow k = \sqrt[3]{{32}} = 2\sqrt[3]{4}.$

**DJ Hobo** · September 3rd 2009, 02:22 AM

Thanks for the help. There are a few things I don't understand, however..

Originally Posted by DeMath

$ax = 4x - {x^2} \Leftrightarrow {x^2} - \left( {4 - a} \right)x = 0 \Leftrightarrow \left[ \begin{gathered} {x_1} = 0, \hfill \\ {x_2} = 4 - a. \hfill \\ \end{gathered} \right.$

Consequently, you got that $k = 4 - a \Leftrightarrow a = 4 - k$

How did you figure out those two x-values? Well, I understand how you got x = 0, but I don't understand how you came to the second value. Also, what do you mean by the "k = ...."

I understand everything else in your solution, just not those two things.

**chug1** · September 3rd 2009, 02:26 AM

For question 12, what they are asking for is the area enclosed by the outline. The first part of this is two semicircles (ie one whole circle) with radius=1. Use the formula for the area of a circle to get this area. The second part of the area is that under a quadratic. See the attached image for the quadratic they mean. The shaded area is one of the two 'pointy' sections of the brooch. Use integrals to find the area under this curve and above the x-axis from x=0 to x=5, multiply it by 2 (for both halves of the right-hand part of the brooch and add it to the circular area found earlier. This should be the area of the brooch.

p.s. sorry about the poor quality of the image, hopefully it's helpful.

**DJ Hobo** · September 3rd 2009, 02:36 AM

Originally Posted by chug1

For question 12, what they are asking for is the area enclosed by the outline. The first part of this is two semicircles (ie one whole circle) with radius=1. Use the formula for the area of a circle to get this area. The second part of the area is that under a quadratic. See the attached image for the quadratic they mean. The shaded area is one of the two 'pointy' sections of the brooch. Use integrals to find the area under this curve and above the x-axis from x=0 to x=5, multiply it by 2 (for both halves of the right-hand part of the brooch and add it to the circular area found earlier. This should be the area of the brooch.

p.s. sorry about the poor quality of the image, hopefully it's helpful.

The image quality is fine, and I understand what you did. Thank you very much for the help.

**DeMath** · September 3rd 2009, 03:02 AM

Originally Posted by DJ Hobo

Thanks for the help. There are a few things I don't understand, however..

How did you figure out those two x-values? Well, I understand how you got x = 0, but I don't understand how you came to the second value. Also, what do you mean by the "k = ...."

I understand everything else in your solution, just not those two things.

In fact, the abscissa $k$ is the upper limit of integration.

See a graphical interpretation of the problem once

**DJ Hobo** · September 3rd 2009, 03:06 AM

Originally Posted by DeMath

In fact, the abscissa $k$ is the upper limit of integration.

See a graphical interpretation of the problem once

I see.. Your diagram does make the question clearer, but I still don't understand how you got "4 - a".

EDIT - I finally understand. Thanks for the help

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