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!

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