(Integration), where did I go wrong? (3 Viewers)

[InteGrand]

So basically, this represents the 'physical area' between the part of the curve below x-axis, and the x-axis: (minus sign is in front because without it, we're not getting the physical area, but a signed area; it'd only be equal to the physical area if the curve was above or on the x-axis for the entirety interval of that integral).

This represents the physical area between the curve from 0 to 2 and the x-axis: (no minus sign needed here, since the curve is never below the x-axis in this interval).

Calculating both these integrals and adding the answers will get you the total physical area.

 

[BlueGas]

Another question, question 1 a) (from the picture above) asks to find the area bounded by the graph, with ordinates x = 2 and x = 3, you can't split these up into sub-intervals right?

 

[BlueGas]

Another question, question 1 a) (from the picture above) asks to find the area bounded by the graph, with ordinates x = 2 and x = 3, you can't split these up into sub-intervals right?

Not sure what's up with Bored of Studies right now it's playing up on me, but if anyone can help me, much appreciated.

 

[FrankXie]

Another question, question 1 a) (from the picture above) asks to find the area bounded by the graph, with ordinates x = 2 and x = 3, you can't split these up into sub-intervals right?

You can always split up. But for this one you do not need to split up because the whole region is above x-axis.

 

[BlueGas]

You can always split up. But for this one you do not need to split up because the whole region is above x-axis.

How would you split it up? Would you do 2 and 2 1/2 and 2 1/2 and 3? Or you can't do that?

 

[BlueGas]

You can always split up. But for this one you do not need to split up because the whole region is above x-axis.

Also what sort of question you would know that part of the region is below the x -axis? I mean that you might come across a graph that has a region below and above the x-axis, what would the question and ordinates be like?

 

[InteGrand]

How would you split it up? Would you do 2 and 2 1/2 and 2 1/2 and 3? Or you can't do that?

If the curve is above the x-axis for the entire interval, you don't need to split it up at all to find the physical area. But you can split it up in any way you want and you'll get the same answer (it'll just take longer so it'll be a waste of time).

 

[InteGrand]

Also what sort of question you would know that part of the region is below the x -axis? I mean that you might come across a graph that has a region below and above the x-axis, what would the question and ordinates be like?

You did one of these yesterday, the function was , and area between the curve and the x-axis was asked for between and . To find the parts below and above the x-axis, you sketch the curve.

 

[BlueGas]

You did one of these yesterday, the function was , and area between the curve and the x-axis was asked for between and . To find the parts below and above the x-axis, you sketch the curve.

This might be a stupid question but how do you know that there is part of the graph below and above the x-axis?

 

[BlueGas]

I just found out a way on how to do these sort of questions with the same steps, here are the pictures:

 

[BlueGas]

So is there anything wrong with my method?

 

[BlueGas]

I just done a different question using this method and it failed. So frustrated right now.

 

[FrankXie]

you never drew a graph? remember, the most important thing of all when comuputing an area is to sketch the graph. From the graph, you would know which part is above or below the x-axis.

 

[BlueGas]

you never drew a graph? remember, the most important thing of all when comuputing an area is to sketch the graph. From the graph, you would know which part is above or below the x-axis.

That's the problem I don't know how to draw a graph, and it's killing me.