The integral of 0 to 1 of √x^2 - 2x +1 dx is?
*The whole equation is under the square root*
So far, I only learned how to do this with the graphing method, and I got stuck when I had to find the area of the curve of the parabola. How do you find the area of a curve? Or am I even doing it right? Please show as much work as possible so I can understand. Thanks in advance. :)
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... ∫ √(x² - 2x + 1) dx ...... on [ 0, 1 ]
= ∫ √(x-1)² dx
= ∫ | x-1 | dx
= (1/2) [ (x-1)·| x-1 | ] ... on [ 0, 1 ]
= (1/2) [ 0·|0| - (-1)·|-1| ]
= (1/2)· [ 0 + 1(1) ]
= (1/2) ........................................................ Ans.
_______________________________
x^2 - 2x + 1 = (x - 1)^2
sqrt((x - 1)^2) * dx =>
(x - 1) * dx
Now integrate
(1/2) * x^2 - x + C
From 0 to 1
(1/2) * (1^2 - 0^2) - (1 - 0) + C - C =>
(1/2) * (1 - 0) - (1) =>
1/2 - 1 =>
-1/2
Thats all honey
x^2 - 2x + 1 = (x - 1)^2
sqrt((x - 1)^2) * dx =>
(x - 1) * dx
Now integrate
(1/2) * x^2 - x + C
From 0 to 1
(1/2) * (1^2 - 0^2) - (1 - 0) + C - C =>
(1/2) * (1 - 0) - (1) =>
1/2 - 1 =>
-1/2