# Evaluate the integral without using tables. \int_{0}^{1}(-\ln x)dx

Evaluate the integral without using tables.
$$\displaystyle{\int_{{{0}}}^{{{1}}}}{\left(-{\ln{{x}}}\right)}{\left.{d}{x}\right.}$$

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diskusje5
Evaluate the integrals as follows:
Substitute $$\displaystyle{u}={\ln{{x}}},{d}{u}={\frac{{{1}}}{{{x}}}}{\left.{d}{x}\right.},{v}'={1},{v}={x}\ \in\ \int{u}{d}{v}={u}{v}-\int{v}{d}{u}$$.
$$\displaystyle{\int_{{{0}}}^{{{1}}}}-{\ln{{\left({x}\right)}}}{\left.{d}{x}\right.}=-{\int_{{{0}}}^{{{1}}}}{\ln{{\left({x}\right)}}}{\left.{d}{x}\right.}$$
$$\displaystyle=-{{\left[{x}{\ln{{\left({x}\right)}}}-\int{1}{\left.{d}{x}\right.}\right]}_{{{0}}}^{{{1}}}}$$
$$\displaystyle=-{{\left[{x}{\ln{{\left({x}\right)}}}-{x}\right]}_{{{0}}}^{{{1}}}}$$
=1