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{e ^{x^{2}} - 1}{x}

n \to \infty lim (- n^{2})

\frac{d}{d x} (2 x^{3} (1 - x^{2})^{4})

x y^{^{'}} - 2 y + 8 = 0

\int e^{4 x + 5} d x

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