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

The integral from-1 to 1 of (y^2-2)-(e^y) is -e+1/e-10/3

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

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

- e + \frac{1}{e} - \frac{10}{3}

Decimal Notation

- 5.68373 \dots

Solution steps

\int_{- 1}^{1} y^{2} - 2 - 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_{- 1}^{1} y^{2} d y - \int_{- 1}^{1} 2 d y - \int_{- 1}^{1} e^{y} d y

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

\int_{- 1}^{1} 2 d y = 4

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

= \frac{2}{3} - 4 - (e - \frac{1}{e})

Simplify

\frac{2}{3} - 4 - (e - \frac{1}{e}) : - e + \frac{1}{e} - \frac{10}{3}

= - e + \frac{1}{e} - \frac{10}{3}

\frac{d}{d x} (x^{3} + 2 y)

x \to \infty lim (t^{x})

x \to 4 lim (x^{2} - 4 x)

x \to - 5 lim (\frac{x + 1}{x + 5})

f (y) = y

What is the integral from-1 to 1 of (y^2-2)-(e^y) ?
The integral from-1 to 1 of (y^2-2)-(e^y) is -e+1/e-10/3