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

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

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integral from 1 to e of ln(y)

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

1

Solution steps

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

Apply Integration By Parts

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

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

\int 1 d y = y

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

Compute the boundaries:

1

= 1

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

\int_{1}^{3} 5^{2 x} d x

\int_{- 1}^{0} - 5 x (x^{2} - 1)^{4} d x

\int_{0}^{1} (x + 8 1 - x^{2}) d x

\int_{1}^{32} 5 x d x

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