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

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

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(B) To evaluate the integral $I = \int_1^e \log x \, dx$,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:$I = [x \log x]_1^e - \int_1^e x \cdot \frac{1}{x} \, dx$$I = [x \log x]_1^e - \int_1^e 1 \, dx$$I = [x \log x - x]_1^e$Now,substitute the limits:$I = (e \log e - e) - (1 \log 1 - 1)$Since $\log e = 1$ and $\log 1 = 0$:$I = (e(1) - e) - (0 - 1)$$I = (e - e) - (-1)$$I = 0 + 1 = 1$.

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

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