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

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}

\int_{0}^{t} (t - x) sin (x) d x

y^{^{'}} + \frac{2}{x} y = 6 \frac{y ^{3}}{x ^{2}}

f (x) = 5 x (1 + x^{2})

3 e^{x} x

h \to + 0 - lim (\frac{2 ^{h} - 1}{h})

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