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

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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 Notation

5.68373 \dots

Solution steps

\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}

\frac{\partial}{\partial y} (tan (3 x - 2 y + 1))

\frac{s + 7}{s ^{2} + 8 s + 25}

g (x) = (3 - 4 x)^{3}

\frac{d}{d x} (\frac{10 x}{x ^{2} + 4})

x \to 1 - lim (\frac{2 x}{ln ( 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