int_0^1 (0.001)^{frac{x}{3}} e^x , dx = | vedclass

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(C) We have the integral $I = \int_0^1 (0.001)^{\frac{x}{3}} e^x \, dx$. First,simplify the term $(0.001)^{\frac{x}{3}}$. Since $0.001 = 10^{-3}$,we h

Vedclass · Accepted Answer

$\frac{e-10}{10(1-\log_e 10)}$

Vedclass · Answer

(C) We have the integral $I = \int_0^1 (0.001)^{\frac{x}{3}} e^x \, dx$. First,simplify the term $(0.001)^{\frac{x}{3}}$. Since $0.001 = 10^{-3}$,we have $(10^{-3})^{\frac{x}{3}} = 10^{-x} = \frac{1}{10^x}$. Thus,the integral becomes $I = \int_0^1 \frac{e^x}{10^x} \, dx = \int_0^1 \left(\frac{e}{10}\right)^x \, dx$. Using the formula $\int a^x \, dx = \frac{a^x}{\ln a} + C$,we get: $I = \left[ \frac{(\frac{e}{10})^x}{\ln(\frac{e}{10})} \right]_0^1 = \frac{\frac{e}{10} - 1}{\ln e - \ln 10} = \frac{\frac{e-10}{10}}{1 - \ln 10}$. Therefore,$I = \frac{e-10}{10(1-\ln 10)}$. This matches option $C$.

$x$	$1$	$2$	$3$	$4$
$y$	$0.7111$	$0.7222$	$0.7333$	$0.7444$

$t \ (\text{in second})$	$0$	$2$	$4$	$6$	$8$	$10$
$v \ (\text{in m/s})$	$0$	$12$	$16$	$20$	$35$	$60$

$\int_0^1 (0.001)^{\frac{x}{3}} e^x \, dx =$

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