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

[InteGrand]

In general, to find the area bounded by a curve and the x-axis between two given x-values, you need to separate this interval into sub-intervals where the curve is below the x-axis (there may be more than one such sub-intervals), and sub-intervals where the curve is above the x-axis. You need to integrate these sub-intervals separately. Integrating over sub-intervals where the function is negative (i.e. below the x-axis) gives you negative answers, so if you're asked for area (a positive thing), you need to flip the sign for answers where you've integrated over a place where the function is negative.

Then you need to add up all the answers to get the total area.

 

[InteGrand]

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

I didn't actually get 2/3, I just integrated from x = ½ to 1, and then was going to double my answer in the post (since I knew the integral over the entire interval was 0, meaning the part below the x-axis had equal area to that above the x-axis), but forgot to do the doubling when answering in the post.

And the curve is negative from x = 0 to ½, so the way to split the integral is
.

So you didn't take the complete negative region in the integral with the integrand with the absolute value sign, just part of it.

 

[BlueGas]

I didn't actually get 2/3, I just integrated from x = ½ to 1, and then was going to double my answer in the post (since I knew the integral over the entire interval was 0, meaning the part below the x-axis had equal area to that above the x-axis), but forgot to do the doubling when answering in the post.

And the curve is negative from x = 0 to ½, so the way to split the integral is
.

So you didn't take the complete negative region in the integral with the integrand with the absolute value sign, just part of it.

How did you know?

 

[InteGrand]

How did you know?

Since you guys did it and said it was 0 (and then I did it to confirm).

But if I didn't know it, I would've separated it into sub-intervals and added them after flipping signs where appropriate.

 

[BlueGas]

Since you guys did it and said it was 0 (and then I did it to confirm).

But if I didn't know it, I would've separated it into sub-intervals and added them after flipping signs where appropriate.

So basically if you get 0 that's when you know have to separate it into sub-intervals?

 

[InteGrand]

So basically if you get 0 that's when you know have to separate it into sub-intervals?

It's possible that you get a positive number by integrating over the entire interval, and that you still need to separate into sub-intervals. This will happen if there are regions below the x-axis, and also regions above it, but the ones above it have greater area than the ones below it, so that the signed area (which integrating over the whole interval gives you) is still positive.

You should just sketch it every time. In this case, it was clear without a sketch that there'd be a negative region, since has a y-intercept of -1 (i.e. when x = 0, which is in the interval in question), which is negative.

 

[InteGrand]

Only if it's obvious that the graph doesn't cross the x-axis in the interval in question, should you not sketch it. e.g. if the function was , you know it is always positive (i.e. above the x-axis), so you won't need to divide into sub-intervals, so to find the area under that curve between two given x-values, just integrate it between those two endpoints.

 

[snowflakeptical]

I didn't actually get 2/3, I just integrated from x = ½ to 1, and then was going to double my answer in the post (since I knew the integral over the entire interval was 0, meaning the part below the x-axis had equal area to that above the x-axis), but forgot to do the doubling when answering in the post.

And the curve is negative from x = 0 to ½, so the way to split the integral is
.

So you didn't take the complete negative region in the integral with the integrand with the absolute value sign, just part of it.

oh so you didn't use 0 as an x value
did you just talk about yourself in 3rd person?

So basically if you get 0 that's when you know have to separate it into sub-intervals?

yeah or when you sketch it, you see the area below the x-axis or when using a value that's negative
sketch it to be sure

 

[InteGrand]

And if the function is clearly negative for the entire given interval, and they ask for the area between the curve and the x-axis, you integrate the function, and then take the negative (or equivalently integrate the negative of the function).

 

[InteGrand]

oh so you didn't use 0 as an x value
did you just talk about yourself in 3rd person?



yeah or when you sketch it, you see the area below the x-axis or when using a value that's negative
sketch it to be sure

Where did I use third person?

And yes, IF integrating over the entire interval gives 0, you'll need to divide into sub-intervals (unless the function is the trivial zero polynomial, f(x) = 0), but if you don't get 0, you may or may not need to, so it's usually a waste of time for these Q's to integrate over the entire interval unless you're sure the graph doesn't cross the x-axis in the interval. Best to sketch it in general.

 

[BlueGas]

Only if it's obvious that the graph doesn't cross the x-axis in the interval in question, should you not sketch it. e.g. if the function was , you know it is always positive (i.e. above the x-axis), so you won't need to divide into sub-intervals, so to find the area under that curve between two given x-values, just integrate it between those two endpoints.

Well that question is done, just one more last question, how would you do Question 4. a) and b)?

 

[InteGrand]

oh so you didn't use 0 as an x value
did you just talk about yourself in 3rd person?

No, I was talking about your integral that had absolute value signs in the integrand.

The integrand is basically the function being integrated.

 

[BlueGas]

Well that question is done, just one more last question, how would you do Question 4. a) and b)?

I already know how to evaluate part a) but I don't know how to do b)

 

[InteGrand]

Well that question is done, just one more last question, how would you do Question 4. a) and b)?

4 a) is just asking to evaluate an integral, not asking for an area, so just integrate it normally.

For 4 b), if you sketch the curve, you'll see it's negative, so you'll need to split it up and integrate the sub-intervals separately (since it's asking for an integral).

If your answer to 4 a) is 0 (I haven't done it), then you can just integrate the positive part (where the function is above the x-axis) and double this.

 

[BlueGas]

4 a) is just asking to evaluate an integral, not asking for an area, so just integrate it normally.

For 4 b), if you sketch the curve, you'll see it's negative, so you'll need to split it up and integrate the sub-intervals separately (since it's asking for an integral).

If your answer to 4 a) is 0 (I haven't done it), then you can just integrate the positive part (where the function is above the x-axis) and double this.

Answer to 4 a) isn't zero, but if you were to split up and integrate the sub-intervals, would the intervals be 0 and -1 and 0 and 2?

 

[InteGrand]

Answer to 4 a) isn't zero, but if you were to split up and integrate the sub-intervals, would the intervals be 0 and -1 and 0 and 2?

Yes

 

[BlueGas]

Yes

Nice, okay, thanks man.

 

[InteGrand]

no probs

 

[BlueGas]

Yes

Will it matter if I do as the intervals -1 and 1 and 1 and 2?

 

[InteGrand]

Will it matter if I do as the intervals -1 and 1 and 1 and 2?

Yes. If you evaluate the integral between -1 and 1, that gives you the net area between -1 and 1 (which is 0). But the question's not asking for net areas, it's asking for the total area, like if you painted the region in question, what area would be painted?

So you'd need to split it up between -1 and 0, and then 0 to 2. Integrating from -1 to 0 gives the net area as -(something), so the actual 'physical area' is something square unit. That's the area of the part below the x-axis.

If you integrated from -1 to 1, for the part below the x-axis (-1 to 0), you won't be evaluating this explicitly, which is what's needed for the question.