What is #f(x) = int (x-1)^3 dx# if #f(-1) = 1 #?

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Answer 1 · 2017-01-03T20:26:07

#f(x) = 1/4(x- 1)^4 - 3#

Let #u = x - 1#. Then #du = dx#.

#f(x) = int u^3 du#

#f(x) = 1/4u^4 + C#

#f(x) = 1/4(x- 1)^4 + C#

Now, solve for #C# knowing that when #x= -1#, #y = 1#.

#1 = 1/4(-1 - 1)^4 + C#

#1 = 1/4(16) + C#

#1 = 4 + C#

#C = -3#

Hopefully this helps!