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sha0403
Group Title
help me to solve this question... Find the total area between the curve y=1x^2, and the xaxis over the interval (0,2)...
 one year ago
 one year ago
sha0403 Group Title
help me to solve this question... Find the total area between the curve y=1x^2, and the xaxis over the interval (0,2)...
 one year ago
 one year ago

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AbhimanyuPudi Group TitleBest ResponseYou've already chosen the best response.0
you have to integrate the given curve's equation in the interval (0,2) \[\int\limits_{0}^{2}1x^2 = (20) + (8/3 + 0) = 2  8/3 = 2/3\] Since are cannot be negative, area = 2/3 sq.units
 one year ago

AbhimanyuPudi Group TitleBest ResponseYou've already chosen the best response.0
sorry I'm little confused while typing equations on OpenStudy..so I've been a little late in answering
 one year ago

sha0403 Group TitleBest ResponseYou've already chosen the best response.0
oo its ok..the working just like that AbhimanyuPudi ?
 one year ago

AbhimanyuPudi Group TitleBest ResponseYou've already chosen the best response.0
may be u have to put another step in between..showing the integration.. \[\int\limits_{0}^{2}1x^2 = x(from 0 \to 2)  x^3/3 (from 0 \to 2)\]
 one year ago

AbhimanyuPudi Group TitleBest ResponseYou've already chosen the best response.0
the negative sign in the answer indicates that the area is formed below the xaxis
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.4
for this case, first u have to figure out of the function y=1x^2 dw:1355817094537:dw so, the total area = A1 + A2
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
i dont think you'll have to consider case of negative area here separately , so integrating directly from 0 to 2 should be correct..i may be wrong though..
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
forexample, integral(cosx) from x=0 to pi = 0 ..ve and +ve areas cancel..
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.4
the result will be different if u use integration (0,2) with i said above :)
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
i know, i am saying if we integrate directly from 0 to 2, we will get correct ans.
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.4
no, @shubhamsrg for A1 = [x1/3*x^3] [0,1] = 2/3 for A2 = [1/3*x^31] [1,2] = 4/3 so, total = 2/3 + 4/3 = 6/3 = 2
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
why will you treat negative area separately ? area can be ve during integration!
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.4
because the rule, if u want calculate area in under xaxist, u have to give ve in front integral (because the integration will be ve also), so we get positive area
 one year ago

sha0403 Group TitleBest ResponseYou've already chosen the best response.0
ok2 which one i should follow?
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.4
btw, for A1 i take int (f(x)) dx [0,1] and A2 i take int(f(x)) dx [1,2]
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
so if ask you integral(cos x) from 0 to pi? ans according to you should be 2 right ?
 one year ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
I absolutely agree with @RadEn as we have to check the position of the area :)
 one year ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
Yes, A = 2 unit square!
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.4
but Abhimanyu's job not 2, but 2/3 :)
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
i just wanted to confirm : http://www.wolframalpha.com/input/?i=integral+%28cosx%29+from+0+to+pi
 one year ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
@sha0403 I leave the computing part for you, questions?
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
yes i have a question// why is wolfram giving value of integral =0 ? ve is also real no.,, ve area we do study, ve indicates direction..please someone clarify..
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.4
@shubhamsrg , is it possible the area be 0 :)
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=integral+%28cosx%29+from+x%3D+pi%2F2+to+3pi%2F2
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
and also, it'll be 0 when you add together the +ve and ve areas!
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.4
if just processing integration, yeah that's right but this case to find the area, so impossible be 0 or negative
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
i see what you mean,,hmm.. now i am also thinking if we go by my methodology , to find area of a circle, suppose x^2 + y^2 =1 , area turns out to be 0 i see..hmm you were right,,we got the take the absolute values..my apologies..
 one year ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
@shubhamsrg I'm not a fan of worf, honestly it's just a simple calculation!
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
i have maybe understood the concept sir,,sorry for the confusion..
 one year ago

sha0403 Group TitleBest ResponseYou've already chosen the best response.0
ok thanks a lot for you guys for help me... =)
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.4
very welcome ... ::)
 one year ago
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