(Integration), where did I go wrong? (1 Viewer)

[BlueGas]

The question is worded very poorly and incorrect, skip it.

Okay, will do.

By the way, the time now is 8:30PM, as you can see from the first post I basically started revising my work form 6:30PM, so 2 hours on this work, do you think the time used was worth it?

 

[hypermax]

im getting 2/3, whats the answer

 

[InteGrand]

The question is worded very poorly and incorrect, skip it.

You shouldn't skip Q.13 in my opinion. It's pretty obvious what it's asking for, it's asking for the area bounded by the curve , the x-axis, and the lines and .

 

[InteGrand]

0 isn't the answer though, that's what confuses me.

An area isn't going to be 0. I'm guessing you just integrated the function from to 1? Remember, this integral gives you the signed area. So parts of the graph below the x-axis will contribute a negative number to the area.

So to do this Q, you'll need to see which part of the curve is below the x-axis, compute the integral in separate regions, and take the negative of the part that's under the x-axis, then add these two together to get the total area.

 

[hypermax]

You shouldn't skip Q.13 in my opinion. It's pretty obvious what it's asking for, it's asking for the area bounded by the curve , the x-axis, and the lines and .

Is it asking for two things in this question

 

[InteGrand]

Okay, will do.

By the way, the time now is 8:30PM, as you can see from the first post I basically started revising my work form 6:30PM, so 2 hours on this work, do you think the time used was worth it?

Well, if you come out of it understanding integration a lot better, then yes, it probably was worth it.

 

[BlueGas]

An area isn't going to be 0. I'm guessing you just integrated the function from to 1? Remember, this integral gives you the signed area. So parts of the graph below the x-axis will contribute a negative number to the area.

So to do this Q, you'll need to see which part of the curve is below the x-axis, compute the integral in separate regions, and take the negative of the part that's under the x-axis, then add these two together to get the total area.

How do I know which part of the curve is below the x-axis?

 

[InteGrand]

Is it asking for two things in this question

Isn't it just asking for the (positive) area under the graph from x = -½ to x = 1 of the curve ?

 

[hypermax]

How do I know which part of the curve is below the x-axis?

the y axis up to the origin

 

[InteGrand]

The answer should be square units.

Edit: I mean 4/3 units. 2/3 is the area above the x-axis (which is also that below the x-axis, since the total integral was 0).

 

[hypermax]

Isn't it just asking for the (positive) area under the graph from x = -½ to x = 1 of the curve ?

but x=1 is out of the graph

 

[hypermax]

The answer should be square units.

yeh thats what i got

 

[BlueGas]

The answer should be square units.

The answer is the booklet is actually 1 1/3 units

 

[InteGrand]

How do I know which part of the curve is below the x-axis?

You basically should draw a quick sketch of . Find its zeros and y-intercept and sketch it from x = -½ to 1. Then you can see which part is below the x-axis.

 

[InteGrand]

The answer is the booklet is actually 1 1/3 units

Sorry, you're right, 4/3 units.

I was looking at my answer for half the total area (just the area above the x-axis, which happens to be that of the one below the x-axis, since the total integral is 0).

 

[InteGrand]

but x=1 is out of the graph

What do you mean? The function is defined at x = 1.

So basically this part of the curve is needed: http://www.graphsketch.com/?eqn1_co..._lines=1&line_width=4&image_w=850&image_h=525

 

[hypermax]

yeah i get it now thanks integrand for helping both of us out

 

[BlueGas]

yeah i get it now thanks integrand for helping both of us out

Lol, I still don't get it, Integrand, can you please post steps on how to do this sort of question?

 

[InteGrand]

Lol, I still don't get it, Integrand, can you please post steps on how to do this sort of question?

Basically, once you have that graph sketched, it's asking for the area underneath the x-axis (i.e. area of the region bounded by the graph and the x-axis from -½ to ½ ), PLUS the area under the curve from ½ to 1.

The area under the curve from -½ to ½ is found by finding , and then taking the negative of this, since that integral is giving you a negative answer (integrating a function in a region where it is below the x-axis gives negative of the area).

If you evaluate that integral (which I think you can do), you get -2/3. So taking the negative of that, you get +2/3. This is the area of the region below the x-axis for that curve (i.e. from x = -½ to ½).

Now evaluate the integral of the function from x = ½ to 1 to find the area of the region that's above the x-axis (this integral is positive since the function is above the x-axis in this region, so you don't need to worry about flipping the sign).

Evaluating it, you'll see it's 2/3. So the area of the part of the curve above the x-axis is 2/3.

So the total area from -½ to 1 is:

(area of part under the x-axis) + (area of part above the x-axis) = 2/3 + 2/3 = 4/3.

 

[snowflakeptical]

um integrand, I got 2/3 like you did before, how is it 4/3 now?
where did I stuff up?
My working out