What is the integral from 0 to 1 of e^y(e^{2y}-1) ?

The integral from 0 to 1 of e^y(e^{2y}-1) is (e^3-1)/3-e+1

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integral from 0 to 1 of e^y(e^{2y}-1)

\int_{0}^{1} e^{y} (e^{2 y} - 1) d y

\frac{e ^{3} - 1}{3} - e + 1

Decimal

4.64356 \dots

Solution steps

\int_{0}^{1} e^{y} (e^{2 y} - 1) d y

Expand

e^{y} (e^{2 y} - 1) : e^{3 y} - e^{y}

= \int_{0}^{1} e^{3 y} - e^{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} e^{3 y} d y - \int_{0}^{1} e^{y} d y

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

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

= \frac{e ^{3} - 1}{3} - (e - 1)

Simplify

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

What is the integral from 0 to 1 of e^y(e^{2y}-1) ?
The integral from 0 to 1 of e^y(e^{2y}-1) is (e^3-1)/3-e+1