int_0^1 {(1 - x)^9} dx = | vedclass

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Question

(B) Let $I = \int_0^1 {(1 - x)^9} dx$. Using the substitution method,let $u = 1 - x$,then $du = -dx$,which implies $dx = -du$. When $x = 0$,$u = 1$. W

Vedclass · Accepted Answer

$\frac{1}{10}$

Vedclass · Answer

(B) Let $I = \int_0^1 {(1 - x)^9} dx$. Using the substitution method,let $u = 1 - x$,then $du = -dx$,which implies $dx = -du$. When $x = 0$,$u = 1$. When $x = 1$,$u = 0$. Substituting these into the integral: $I = \int_1^0 u^9 (-du) = \int_0^1 u^9 du$. Evaluating the integral: $I = \left[ \frac{u^{10}}{10} \right]_0^1 = \frac{1^{10}}{10} - \frac{0^{10}}{10} = \frac{1}{10} - 0 = \frac{1}{10}$.

$\int_0^1 {(1 - x)^9} dx = $

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