What is the correct notation for setting up area integrals?

[silicon_hobo]

Homework Statement

I must divide a region into integrals to find the area. I'm not sure how to write it. Do I need to setup 3 separate integrals or can I just multiply the second one by 2 since I know the parabola is symmetric to the x-axis and the area bounded by the parabola and the dotted line above is equal to that which is below. What's the correct convention to set this up? Thanks again, this forum has been very helpful.

Homework Equations

The Attempt at a Solution

http://www.mcp-server.com/~lush/shillmud/integ1.4.JPG

http://www.mcp-server.com/~lush/shillmud/graph1.4.JPG

 

[sutupidmath]

Homework Statement

I must divide a region into integrals to find the area. I'm not sure how to write it. Do I need to setup 3 separate integrals or can I just multiply the second one by 2 since I know the parabola is symmetric to the x-axis and the area bounded by the parabola and the dotted line above is equal to that which is below. What's the correct convention to set this up? Thanks again, this forum has been very helpful.

Homework Equations

The Attempt at a Solution

http://www.mcp-server.com/~lush/shillmud/integ1.4.JPG

http://www.mcp-server.com/~lush/shillmud/graph1.4.JPG

i do not think you have correctly set up the problem

$$A=\int_{-4}^{0}|-2\sqrt(-x)|dx-\int_{-4}^{-2}|(2x+4)|dx+\int_{-2}^{-1}(2x+4)dx+\int_{-1}^{0}2\sqrt(-x)dx$$

I think this is how it should read,if you want to find the area enclosed by the line you said and the parabola!
or also you could not use the absolute values for the first two integrals, but if this is the case, then you should switch the sign before the integrals to make sure that you will get a positive value, since the area cannot be negative.!

 

[silicon_hobo]

Thanks for the reply. As is often the case in these matters, your answer leads me to more questions. Why do you take the absolute value of the first two functions and, in the second half, why have you added instead of subtracting the integral of 2x+4 from that of 2(-x^1/2). Cheers.

 

[sutupidmath]

Thanks for the reply. As is often the case in these matters, your answer leads me to more questions. Why do you take the absolute value of the first two functions and why have you included the integral of the straight line a second time? Is it not possible to split it into just two parts as in my sketch? Cheers.

well, the first integral calculates the area that the parabola defined with the eq.

$$y=-2\sqrt-x$$ encloses from -4 to 0 with the x-axis. I have put the eq under the abs value since the parabola is under the x-axis it is neg. but the area cannot be negative since it has no meaning to say a rectangle has an area of -4m^2, say, if you do not like abs. value sign, then you have to put an extra minus sign in front of that integral, to make sure that the rez. will be positive.
the second integral calculates the area that the line y=2x+4 encloses with the x-axis from -4 to -2. For the same reason i used abs values here.
the third integral calculates the area that the line y=2x+4 and the x-axis enclose from -2 to -1 above the x-axis. This part is positive so no need to use abs values. the fourth integral calculates the area that the parabola $$y=2\sqrt-x$$ and the x-axis enclose(above the x-axis) from -1 to 0.

so in order to get the area we want we have to subtract the area that the line and x-axis(under x-axis, on the neg part) encloses from -4 to -2.

do u get it?

 

[silicon_hobo]

Okay, I think I can dig it. Thanks. Just to confirm: there's no simpler way to set this up involving my dotted line? Is the other answer I've come up with in the mean time totally wrong?

 

[sutupidmath]

Okay, I think I can dig it. Thanks. Just to confirm: there's no simpler way to set this up involving my dotted line? Is the other answer I've come up with in the mean time totally wrong?

the first integral you set up for finding the area that the line closes up with the x-axis from the point -4 to -1 doesn not really make sense, since you have the line crossing the x-axis at -2. SO,in these cases you have to brake the integral into two parts. Remember that
$$\int_{a}^bf(x)dx$$ only gives you the area under the graph of the function f(x).