C2 Quick Integration Question:
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Hey do you still need help on it?
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Posted from TSR Mobile
Original post by bigboateng
Yes please
Original post by creativebuzz
Yes please
I can't remember off the top of my head but can't you just integrate between 1 and 8?
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How far have you got with the problem? Could you please post some of your current working in order to help you further?
Original post by creativebuzz
I've drawn a box around the part I need help on..
when y = 1 what is x and when y = 8 what is x? Does that give you any clues?
Original post by maggiehodgson
when y = 1 what is x and when y = 8 what is x? Does that give you any clues?
When y=1, x=0
and when y=8, x=0
So could you integrate the curve and then minus... I really don't know :/
Using the picture of the graph above, draw the area you found in part (a).
Add this area to a 1 by 8 rectangle, and compare this area to the one you need to find. What must you now do to get the correct area?
Add this area to a 1 by 8 rectangle, and compare this area to the one you need to find. What must you now do to get the correct area?
Original post by WishingChaff
Using the picture of the graph above, draw the area you found in part (a).
Add this area to a 1 by 8 rectangle, and compare this area to the one you need to find. What must you now do to get the correct area?
Add this area to a 1 by 8 rectangle, and compare this area to the one you need to find. What must you now do to get the correct area?
I need to minus the area under 1.. but how do I do that?
You effectively said it in your answer. Take off the rectangle of area 1 by 2. This should hopefully yield the correct answer.
Original post by creativebuzz
I need to minus the area under 1.. but how do I do that?
Original post by creativebuzz
When y=1, x=0
and when y=8, x=0
So could you integrate the curve and then minus... I really don't know :/
and when y=8, x=0
So could you integrate the curve and then minus... I really don't know :/
No, I don't think so. When y =1 x^3 = 8 does it not. Have another look and see if you see anything now.
Original post by WishingChaff
You effectively said it in your answer. Take off the rectangle of area 1 by 2. This should hopefully yield the correct answer.
I don't see how the area under 1 is the shape of a rectangle.. I mean it's right side is curved not straight...
Original post by maggiehodgson
No, I don't think so. When y =1 x^3 = 8 does it not. Have another look and see if you see anything now.
But the picture itself is showing y=8 and y=1 on the y axis..
Original post by creativebuzz
But the picture itself is showing y=8 and y=1 on the y axis..
So you've integrated in part a) and that has given you the area under the curve between those two limits. Yes?
But the area it's asking for is that little bit PLUS the rectangle to the left (and you know how to calculate that as you have the length and the width. And then there's a bit at the bottom that's not needed so find that area and do a subtraction.
I hope I've explained this properly. I haven't got the picture in front of me at the same time as this reply. I'll check it and if you come back to me I'll know I need to explain better.
Original post by creativebuzz
I don't see how the area under 1 is the shape of a rectangle.. I mean it's right side is curved not straight...
You divide the shaded area into 2 parts. One is a 7x1 rectangle, the other is the part you found in a).
(edited 9 years ago)
Original post by creativebuzz
But the picture itself is showing y=8 and y=1 on the y axis..
I'm posting a diagram that might help you.
Original post by maggiehodgson
So you've integrated in part a) and that has given you the area under the curve between those two limits. Yes?
But the area it's asking for is that little bit PLUS the rectangle to the left (and you know how to calculate that as you have the length and the width. And then there's a bit at the bottom that's not needed so find that area and do a subtraction.
I hope I've explained this properly. I haven't got the picture in front of me at the same time as this reply. I'll check it and if you come back to me I'll know I need to explain better.
But the area it's asking for is that little bit PLUS the rectangle to the left (and you know how to calculate that as you have the length and the width. And then there's a bit at the bottom that's not needed so find that area and do a subtraction.
I hope I've explained this properly. I haven't got the picture in front of me at the same time as this reply. I'll check it and if you come back to me I'll know I need to explain better.
Okay, I think I'm starting to grasp it! This is what I'm currently visualising:
I know that the height of the rectangle is 8 (or is it 7?), but how do I find the length?
And if I were to minus the area under 1, surely I can't just find treat it as a rectangle because it's curved (the part I circled in red)!
(Sorry for being so pernickety, I just really want to make sure I've understood every little bit)!
Original post by maggiehodgson
I'm posting a diagram that might help you.
Ah thank you, that really helped me to understand! But how did you find out the 2 on the xaxis?
Original post by creativebuzz
Ah thank you, that really helped me to understand! But how did you find out the 2 on the xaxis?
Because when y = 1, x cubed = 8 (from the original equation of the graph) so x = 2.
Has that sorted you out?
I've attached a diagram to hopefully make this a bit clearer.
Others have incorrectly said to integrate using 8 and 1 - you can only integrate between x-values.
So you need to find the x-values when y=8 and when y=1 and then integrate this. You should realise that you did this in part a). Now, if you look at my diagram you should see how you could split the graph into a rectangle and another area. Working out the area of the rectangle (which is in the shaded region) should be straightfoward, the vertical height is from 1 to 8.
However you must be careful, because when you integrated you calculated the area under the curve to the x-axis, and the shaded region does not include the x-axis. So you need to work out this area (straightforward) and take it away from the integrated area.
Area of rectangle + area of integrated area (with the bit at the bottom taken away) = total area
Others have incorrectly said to integrate using 8 and 1 - you can only integrate between x-values.
So you need to find the x-values when y=8 and when y=1 and then integrate this. You should realise that you did this in part a). Now, if you look at my diagram you should see how you could split the graph into a rectangle and another area. Working out the area of the rectangle (which is in the shaded region) should be straightfoward, the vertical height is from 1 to 8.
However you must be careful, because when you integrated you calculated the area under the curve to the x-axis, and the shaded region does not include the x-axis. So you need to work out this area (straightforward) and take it away from the integrated area.
Area of rectangle + area of integrated area (with the bit at the bottom taken away) = total area
(edited 9 years ago)
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