I'm in highschool and just started integration. I found the following problem in an old question paper and I find it very challenging. ∫ 0 1 x 2 − 2 x + 1 d xso I simplified it algebraically to ∫ 0 1 ( x − 1 ) 2 d xwhich of course is ∫ 0 1 | x − 1 | d xas the absolute value is a linear function over x ∈ [ 0 , ∞ ) so I proceed to evaluate it as x 2 / 2 − x for upper limit 1 and lower limit 0 which is ( ( 1 ) 2 / 2 − 1 ) − ( 0 − 0 ) and equals − 1 2 , but according to wolfram alpha it is 1 2 .Please explain at beginner level.

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I&#039;m in highschool and just started integration. I found the following problem in an old question paper and I find it very challenging.      ∫    0    1              x      2        −    2    x    +    1    d  xso I simplified it algebraically to      ∫    0    1        (    x    −    1          )      2        d  xwhich of course is      ∫    0    1        |    x  −  1      |    d  xas the absolute value is a linear function over   x  ∈  [  0  ,  ∞  ) so I proceed to evaluate it as       x    2        /    2  −  x for upper limit 1 and lower limit 0 which is   (  (  1      )    2        /    2  −  1  )  −  (  0  −  0  ) and equals   −      1    2  , but according to wolfram alpha it is       1    2  .Please explain at beginner level.

vikafa4g · Accepted Answer

For   x  ∈  (  0  ,  1  )  ,      |    x  −  1      |    =  1  −  xHence we get       1    2  .Note that integrating a nonnegative function gives you nonnegative solution.

Monfredo0n · Accepted Answer

Note that when   x  ≤  1 and   x  ≥  0       |    x  −  1      |    =  1  −  x from here you can get your answer of 1/2Your mistake is that you say       |    x  −  1      |   is equal to   x  −  1 in   [  0  ,  ∞  ) this is wrong it is only true for   x  ≥  1 but this does not help you .

I'm in highschool and just started integration. I found the following problem in an old question pap

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