What is the integral from-1 to 1 of (y^2-4)-(e^y) ?

The integral from-1 to 1 of (y^2-4)-(e^y) is -e+1/e-22/3

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integral from-1 to 1 of (y^2-4)-(e^y)

\int_{- 1}^{1} (y^{2} - 4) - (e^{y}) d y

- e + \frac{1}{e} - \frac{22}{3}

Decimal

- 9.68373 \dots

Solution steps

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

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

\int_{- 1}^{1} 4 d y = 8

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

= \frac{2}{3} - 8 - (e - \frac{1}{e})

Simplify

\frac{2}{3} - 8 - (e - \frac{1}{e}) : - e + \frac{1}{e} - \frac{22}{3}

= - e + \frac{1}{e} - \frac{22}{3}

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

\int_{0}^{9} x^{2} - 4 d x

\int_{- 2}^{3} (x + 2) - (x^{2} - 4) d x

\int_{0}^{2} 12 d x

\int_{0}^{5} 2 t e^{- t^{2}} d t

What is the integral from-1 to 1 of (y^2-4)-(e^y) ?
The integral from-1 to 1 of (y^2-4)-(e^y) is -e+1/e-22/3