int x^5 e^{-2 x} d x= | vedclass

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Question

(B) To evaluate $\int x^5 e^{-2 x} d x$,we use the formula for integration by parts repeatedly or the tabular method (Bernoulli's formula): $\int u v'

Vedclass · Accepted Answer

$-e^{-2 x}\left[\frac{x^5}{2}+\frac{5 x^4}{4}+\frac{5 x^3}{2}+\frac{15 x^2}{4}+\frac{15 x}{4}+\frac{15}{8}\right]+c$

Vedclass · Answer

(B) To evaluate $\int x^5 e^{-2 x} d x$,we use the formula for integration by parts repeatedly or the tabular method (Bernoulli's formula): $\int u v' dx = u v - u' v_1 + u'' v_2 - u''' v_3 + \dots$Here,$u = x^5$ and $v' = e^{-2 x}$.$u = x^5, u' = 5x^4, u'' = 20x^3, u''' = 60x^2, u^{(4)} = 120x, u^{(5)} = 120, u^{(6)} = 0$.$v = e^{-2 x}, v_1 = \frac{e^{-2 x}}{-2}, v_2 = \frac{e^{-2 x}}{4}, v_3 = \frac{e^{-2 x}}{-8}, v_4 = \frac{e^{-2 x}}{16}, v_5 = \frac{e^{-2 x}}{-32}, v_6 = \frac{e^{-2 x}}{64}$.Applying the formula:$\int x^5 e^{-2 x} d x = x^5 \left(\frac{e^{-2 x}}{-2}\right) - 5x^4 \left(\frac{e^{-2 x}}{4}\right) + 20x^3 \left(\frac{e^{-2 x}}{-8}\right) - 60x^2 \left(\frac{e^{-2 x}}{16}\right) + 120x \left(\frac{e^{-2 x}}{-32}\right) - 120 \left(\frac{e^{-2 x}}{64}\right) + c$$= -e^{-2 x} \left[ \frac{x^5}{2} + \frac{5x^4}{4} + \frac{20x^3}{8} + \frac{60x^2}{16} + \frac{120x}{32} + \frac{120}{64} \right] + c$$= -e^{-2 x} \left[ \frac{x^5}{2} + \frac{5x^4}{4} + \frac{5x^3}{2} + \frac{15x^2}{4} + \frac{15x}{4} + \frac{15}{8} \right] + c$.

$\int x^5 e^{-2 x} d x=$

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