int_0^2 x^3(2-x)^4 , dx = | vedclass

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(D) Let $I = \int_0^2 x^3(2-x)^4 \, dx$. Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$, we get: $I = \int_0^2 (2-x)^3(2-(2-x))^4 \,

Vedclass · Accepted Answer

$\frac{32}{35}$

Vedclass · Answer

(D) Let $I = \int_0^2 x^3(2-x)^4 \, dx$. Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$, we get: $I = \int_0^2 (2-x)^3(2-(2-x))^4 \, dx = \int_0^2 (2-x)^3 x^4 \, dx$. Expanding the terms: $I = \int_0^2 (8 - 12x + 6x^2 - x^3) x^4 \, dx = \int_0^2 (8x^4 - 12x^5 + 6x^6 - x^7) \, dx$. Integrating term by term: $I = \left[ \frac{8x^5}{5} - \frac{12x^6}{6} + \frac{6x^7}{7} - \frac{x^8}{8} \right]_0^2$. $I = \left[ \frac{8(32)}{5} - 2(64) + \frac{6(128)}{7} - \frac{256}{8} \right]$. $I = \frac{256}{5} - 128 + \frac{768}{7} - 32 = \frac{256}{5} + \frac{768}{7} - 160$. $I = \frac{1792 + 3840 - 5600}{35} = \frac{32}{35}$.

$\int_0^2 x^3(2-x)^4 \, dx = $

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