The value of int_1^2 {log x,dx} is | vedclass

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(C) To evaluate the integral $\int_1^2 \log x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = \log x$ a

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$\log(4/e)$

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(C) To evaluate the integral $\int_1^2 \log x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = \log x$ and $dv = dx$. Then $du = \frac{1}{x} dx$ and $v = x$. Applying the formula: $\int \log x \, dx = x \log x - \int x \cdot \frac{1}{x} \, dx = x \log x - \int 1 \, dx = x \log x - x$. Now,applying the limits from $1$ to $2$: $[x \log x - x]_1^2 = (2 \log 2 - 2) - (1 \log 1 - 1)$. Since $\log 1 = 0$,we have: $(2 \log 2 - 2) - (0 - 1) = 2 \log 2 - 2 + 1 = 2 \log 2 - 1$. Using the property $n \log a = \log(a^n)$,we get $2 \log 2 = \log(2^2) = \log 4$. Also,$1 = \log e$. Therefore,$\log 4 - \log e = \log(4/e)$.

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

The value of $\int_1^2 {\log x\,dx} $ is

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