What is the integral from 1 to e of 1-ln(y) ?

The integral from 1 to e of 1-ln(y) is e-2

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integral from 1 to e of 1-ln(y)

\int_{1}^{e} 1 - ln (y) d y

e - 2

Decimal Notation

0.71828 \dots

Solution steps

\int_{1}^{e} 1 - ln (y) d y

Apply Integration By Parts

: [y (1 - ln (y)) - \int - 1 d y]_{1}^{e}

= [y (1 - ln (y)) - \int - 1 d y]_{1}^{e}

\int - 1 d y = - y

= [y (1 - ln (y)) - (- y)]_{1}^{e}

Simplify

[y (1 - ln (y)) - (- y)]_{1}^{e} : [2 y - y ln (y)]_{1}^{e}

= [2 y - y ln (y)]_{1}^{e}

Compute the boundaries:

e - 2

= e - 2

\int_{0}^{5} (x + 3) d x

\int_{3}^{5} - 5 t d t

\int_{0}^{1} \frac{x e ^{- 2 x}}{2} d x

\int_{- 1}^{1} \frac{x + 2}{x} d x

\int_{a}^{b} ((a x)^{3} + e^{- b x}) d x

What is the integral from 1 to e of 1-ln(y) ?
The integral from 1 to e of 1-ln(y) is e-2