What is the integral from 0 to 1 of (e-1)/(2e) ?

The integral from 0 to 1 of (e-1)/(2e) is (e-1)/(2e)

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integral from 0 to 1 of (e-1)/(2e)

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

\frac{e - 1}{2 e}

Decimal

0.31606 \dots

Solution steps

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

Integral of a constant:

\int a d x = a x

= [\frac{e - 1}{2 e} x]_{0}^{1}

Compute the boundaries:

\frac{e - 1}{2 e}

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

\int_{0}^{1} e^{π x} cos (π) x d x

\int_{1}^{2} \frac{v ^{2} + 3 v ^{5}}{v ^{3}} d v

\int_{1}^{e} 8 x ln (x) d x

\int_{0}^{2} sin (- x^{2}) d x

\int_{1}^{2} - \frac{5}{x ^{2}} d x

What is the integral from 0 to 1 of (e-1)/(2e) ?
The integral from 0 to 1 of (e-1)/(2e) is (e-1)/(2e)