What is the integral from 0 to 1 of y(-1+y+e-e^y) ?

The integral from 0 to 1 of y(-1+y+e-e^y) is -7/6+e/2

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integral from 0 to 1 of y(-1+y+e-e^y)

\int_{0}^{1} y (- 1 + y + e - e^{y}) d y

- \frac{7}{6} + \frac{e}{2}

Decimal

0.19247 \dots

Solution steps

\int_{0}^{1} y (- 1 + y + e - e^{y}) d y

Expand

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

= \int_{0}^{1} - y + y^{2} + ey - e^{y} y d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{0}^{1} y d y + \int_{0}^{1} y^{2} d y + \int_{0}^{1} ey d y - \int_{0}^{1} e^{y} y d y

\int_{0}^{1} y d y = \frac{1}{2}

\int_{0}^{1} y^{2} d y = \frac{1}{3}

\int_{0}^{1} ey d y = \frac{e}{2}

\int_{0}^{1} e^{y} y d y = 1

= - \frac{1}{2} + \frac{1}{3} + \frac{e}{2} - 1

Combine the fractions

- 1 - \frac{1}{2} + \frac{1}{3} : - \frac{7}{6}

= - \frac{7}{6} + \frac{e}{2}

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What is the integral from 0 to 1 of y(-1+y+e-e^y) ?
The integral from 0 to 1 of y(-1+y+e-e^y) is -7/6+e/2