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} in x d x

\frac{( e ^{2} - 1 ) in}{2}

Solution steps

\int_{1}^{e} in x d x

Take the constant out:

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

= in \cdot \int_{1}^{e} x d x

Apply the Power Rule

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

Compute the boundaries:

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

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

Simplify

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

\int_{0}^{1} e^{x} - x \cdot e^{x} d x

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

\int_{0}^{0.5} b - 2 b x d x

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