int_0^2 x^2(2-x)^5 d x= | vedclass

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(C) To evaluate the integral $I = \int_0^2 x^2(2-x)^5 dx$,we use the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$. Substituting $x = 2-t$,we get $

Vedclass · Accepted Answer

$\frac{32}{21}$

Vedclass · Answer

(C) To evaluate the integral $I = \int_0^2 x^2(2-x)^5 dx$,we use the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$. Substituting $x = 2-t$,we get $dx = -dt$. When $x=0, t=2$ and when $x=2, t=0$. $I = \int_2^0 (2-t)^2(t)^5 (-dt) = \int_0^2 (4 - 4t + t^2)t^5 dt$. $I = \int_0^2 (4t^5 - 4t^6 + t^7) dt$. Integrating term by term: $I = [\frac{4t^6}{6} - \frac{4t^7}{7} + \frac{t^8}{8}]_0^2$. $I = [\frac{2(2^6)}{3} - \frac{4(2^7)}{7} + \frac{2^8}{8}] = [\frac{128}{3} - \frac{512}{7} + 32]$. Finding a common denominator of $21$: $I = \frac{896 - 1536 + 672}{21} = \frac{32}{21}$. Thus,the correct option is $C$.

$\int_0^2 x^2(2-x)^5 d x=$

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