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

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

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integral from e to 1 of ln(x)

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

- 1

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Apply Integration By Parts

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

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

\int 1 d x = x

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

Compute the boundaries:

1

= - 1

\int_{0}^{1} 1 + 16 x^{2} d x

\int_{0}^{4} - 2 x + 3 d x

\int_{0}^{2} x^{3} + 1 d x

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

\int_{1}^{4} 4 x d x

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