What is the integral from 1 to ln(8) of e^y(y-1) ?

The integral from 1 to ln(8) of e^y(y-1) is 24ln(2)-16+e

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integral from 1 to ln(8) of e^y(y-1)

\int_{1}^{l n (8)} e^{y} (y - 1) d y

24 ln (2) - 16 + e

Decimal

3.35381 \dots

Solution steps

\int_{1}^{l n (8)} e^{y} (y - 1) d y

Apply Integration By Parts

: [e^{y} (y - 1) - \int e^{y} d y]_{1}^{3 l n (2)}

= [e^{y} (y - 1) - \int e^{y} d y]_{1}^{3 l n (2)}

\int e^{y} d y = e^{y}

= [e^{y} (y - 1) - e^{y}]_{1}^{3 l n (2)}

Expand

e^{y} (y - 1) - e^{y} : e^{y} y - 2 e^{y}

= [e^{y} y - 2 e^{y}]_{1}^{3 l n (2)}

Compute the boundaries:

24 ln (2) - 16 + e

= 24 ln (2) - 16 + e

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What is the integral from 1 to ln(8) of e^y(y-1) ?
The integral from 1 to ln(8) of e^y(y-1) is 24ln(2)-16+e