What is the integral from 1 to e of Inx ?

The integral from 1 to e of Inx is ((e^2-1)In)/2

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integral from 1 to e of Inx

\int_{1}^{e} I n x d x

\frac{( e ^{2} - 1 ) I n}{2}

Solution steps

\int_{1}^{e} I n x d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= I n \cdot \int_{1}^{e} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{1}^{e}

= I n [\frac{x ^{2}}{2}]_{1}^{e}

Compute the boundaries:

\frac{e ^{2}}{2} - \frac{1}{2}

= I n (\frac{e ^{2}}{2} - \frac{1}{2})

Simplify

= \frac{( e ^{2} - 1 ) I n}{2}

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What is the integral from 1 to e of Inx ?
The integral from 1 to e of Inx is ((e^2-1)In)/2