Evaluate the definite integral int_{2}^{3} fr | vedclass

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(B) Let $I = \int_{2}^{3} \frac{1}{x} d x$. We know that the antiderivative of $\frac{1}{x}$ is $\log |x|$. By the second fundamental theorem of calcu

Vedclass · Accepted Answer

$\log \frac{3}{2}$

Vedclass · Answer

(B) Let $I = \int_{2}^{3} \frac{1}{x} d x$. We know that the antiderivative of $\frac{1}{x}$ is $\log |x|$. By the second fundamental theorem of calculus,$\int_{a}^{b} f(x) d x = F(b) - F(a)$,where $F(x)$ is the antiderivative of $f(x)$. Here,$F(x) = \log |x|$. Therefore,$I = F(3) - F(2) = \log |3| - \log |2|$. Using the logarithmic property $\log a - \log b = \log \frac{a}{b}$,we get $I = \log \frac{3}{2}$.

Evaluate the definite integral $\int_{2}^{3} \frac{1}{x} d x$.

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