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

The integral from-1 to 1 of (e^y-(y^2-3)) is e-1/e+16/3

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integral from-1 to 1 of (e^y-(y^2-3))

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

e - \frac{1}{e} + \frac{16}{3}

Decimal Notation

7.68373 \dots

Solution steps

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

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

\int_{- 1}^{1} y^{2} - 3 d y = - \frac{16}{3}

= e - \frac{1}{e} - (- \frac{16}{3})

Simplify

= e - \frac{1}{e} + \frac{16}{3}

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What is the integral from-1 to 1 of (e^y-(y^2-3)) ?
The integral from-1 to 1 of (e^y-(y^2-3)) is e-1/e+16/3