Find the integral of #(x-10)^2#?

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Answer 1 · 2017-04-02T15:59:16

#int(x-10)^2dx=(x-10)^3/3+c#

We have to find #int(x-10)^2dx#

let #x-10=u# then #dx=du# and

#int(x-10)^2dx#

= #intu^2du#

= #u^3/3+c#

= #(x-10)^3/3+c#

Answer 2 · 2017-04-02T16:02:25

Please see below.

Method 1

#(x-10)^2 = x^2-20x+100#, so

#int (x-10)^2 dx = int (x^2-20x+100) dx#

# = x^3/3-10x^2+100x +C#

Method 2

Let #u = x-10#. The results in #du = 1 dx#. Change the integral

#int underbrace((x-10))_u^2 underbrace(dx)_(du) = int u^2 du#

# = u^3/3 +C_2#. Now undo the substitution to get

# = (x-10)^3/3 +C_2#

Note

Expanding the second answer, we get

#(x-10)^3/3 +C = (x^3-30x^2+300x-1000)/3 +C_2#

# = x^3/3-10x^2+100x-1000/3 +C_2#

So the two solution methods lead to different constants.

#C = C_2-1000/3#