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hello, my problem is that when i try to do it i open it to x^3-2x and when i do integration it becomes x^4/4-x^2 which is wrong according to the answer given by the calculator which is (x^2-2)^2/4

 

help please! thanks!

 Oct 23, 2015

Best Answer 

 #1
avatar+27475 
+10

If you expand (x^2-2)^2/4 you get x^4/4 - x^2 + 1

 

Although this looks different from x^4/4 - x^2 you have to remember that indefinite integrals are only defined to within a constant, so really, doing the integration your way you should have had x^2/4 - x^2 + k1 and the calculator should have had (x^2 - 2)^2/4 + k2 where k1 and k2 are (different) constants, related by k1 = k2+1.

 Oct 23, 2015
 #1
avatar+27475 
+10
Best Answer

If you expand (x^2-2)^2/4 you get x^4/4 - x^2 + 1

 

Although this looks different from x^4/4 - x^2 you have to remember that indefinite integrals are only defined to within a constant, so really, doing the integration your way you should have had x^2/4 - x^2 + k1 and the calculator should have had (x^2 - 2)^2/4 + k2 where k1 and k2 are (different) constants, related by k1 = k2+1.

Alan Oct 23, 2015

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