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(x^2 + y^2) = 2(x^2 – y^2)How do i solve this?
What exactly are vertical and horizontal elements?Need help with the 6th question. Could somebody please explain this?View attachment 57703
...you can't answer me if you don't knowWhat exactly are vertical and horizontal elements?
Limits?
I know everything about integration but ive never seen 'vertical or horizontal elements' in any of the questions...you can't answer me if you don't know
It's part of Integration concept...horizontal element is a strip going horizontally, the area of which we have gotta find using integration, same goes for vertical elements. The main thing is we can find the area under a curve on a graph, by finding integrals of the equation of curve.
If you've covered integration already, you'd know!
can u tell me the answer.?...you can't answer me if you don't know
It's part of Integration concept...horizontal element is a strip going horizontally, the area of which we have gotta find using integration, same goes for vertical elements. The main thing is we can find the area under a curve on a graph, by finding integrals of the equation of curve.
If you've covered integration already, you'd know!
Accurate! very nice indeed. You got it right for the first part! but not for the second though, it's the same as first both parts have answer 42.67can u tell me the answer.?
im getting 42.67 for the first part. and 0.5 for the second
so how did you do it? could you please show me the procedure and explain a little?can u tell me the answer.?
im getting 42.67 for the first part. and 0.5 for the second
oh yes ! *2 sec dance party*Accurate! very nice indeed. You got it right for the first part! but not for the second though, it's the same as first both parts have answer 42.67
sure. give me 5 minutes to type itso how did you do it? could you please show me the procedure and explain a little?
lettme join too!oh yes ! *2 sec dance party*
Your concepts are clear. No doubt. Since you know integration. In this question you are not given the limits but youre told that you have to find the area in the first quadrant. This suggests that there is no negative limit. So if you plug in a few values of x in the equation you will get x as 4 when y equals 0. and when x is 0 you get y as 16. Therefore your limits are 4 and 0.so how did you do it? could you please show me the procedure and explain a little?
*joins*lettme join too!
You can even make a quick sketch of the graphy. Thats what I did. Then it was clear that 4 and 0 are the required limitsso how did you do it? could you please show me the procedure and explain a little?
yah...I totally get it! ...thank you so muchYour concepts are clear. No doubt. Since you know integration. In this question you are not given the limits but youre told that you have to find the area in the first quadrant. This suggests that there is no negative limit. So if you plug in a few values of x in the equation you will get x as 4 when y equals 0. and when x is 0 you get y as 16. Therefore your limits are 4 and 0.
Now just simply integrate the equation. You will get 16x - x^3/3. Plug in the limits and youll get 42.67 I hope that explain for you
Epic fun!*joins*
That was funnnn. No?
yah...I totally get it! ...thank you so much
Actually, I did make a sketch, and found one limit as 4 and since the other (-4) wasn't in the first quadrant, I was kinda confused, but now I get it!
Hahaha!Epic fun!
Are you A'Level student?Hahaha!
Was Gave my papers this novemberAre you A'Level student?
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