# A few Integral problems

• Sep 3rd 2009, 12:01 AM
DJ Hobo
A few Integral problems
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. (Nod)
• Sep 3rd 2009, 02:06 AM
DeMath
Quote:

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. (Nod)

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 $k = 4 - a \Leftrightarrow a = 4 - k$

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

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

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

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

Finally you have

$\frac{{{k^3}}}{6} = \frac{1}{2} \cdot \frac{{32}}{3} \Leftrightarrow {k^3} = 32 \Rightarrow k = \sqrt[3]{{32}} = 2\sqrt[3]{4}.$
• Sep 3rd 2009, 02:22 AM
DJ Hobo
Thanks for the help. There are a few things I don't understand, however..
Quote:

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.
• Sep 3rd 2009, 02:26 AM
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.
• Sep 3rd 2009, 02:36 AM
DJ Hobo
Quote:

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.
• Sep 3rd 2009, 03:02 AM
DeMath
Quote:

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

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