How do you find the antiderivative of #int (1-t)^9dt#?

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Answer 1 · 2016-11-13T15:36:47

The integral is #y =-1/10(t-1)^10 + C#.

Let #u = 1 - t#. Then #du = -1(dt) -> dt= (du)/-1 = -(du)#.

#=>-int(u^9)du#

We can now use the rule #int(x^n)dx = (x^(n + 1))/(n + 1) + C#.

#=>-1/10u^10 + C#

Recall that #u = 1 - t#.

#=>-1/10(t - 1)^10 + C#

Hopefully this helps!