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

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

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

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

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

Decimal

5.68373 \dots

Solution steps

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

Simplify

= \int_{- 1}^{1} e^{y} - y^{2} + 2 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} d y + \int_{- 1}^{1} 2 d y

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

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

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

= e - \frac{1}{e} - \frac{2}{3} + 4

Combine the fractions

4 - \frac{2}{3} : \frac{10}{3}

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

n = 2 \sum \infty \frac{3}{4 ^{n}}

\frac{\partial}{\partial x} (ln (1 + x^{2} + y^{2}))

\frac{\partial}{\partial x} (13.12 + 0.6215 x - 11.37 y^{0.16} + 0.3965 x y^{0.16})

t an g e n t f (x) = sin (x) cos (x)

\int \frac{x}{( x ^{2} + 1 ) ( x ^{2} + 4 x + 4 )} d x

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