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

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

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

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

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

Decimal Notation

5.68373 \dots

Solution steps

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

Simplify

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

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^{x} d x - \int_{- 1}^{1} x^{2} d x + \int_{- 1}^{1} 2 d x

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

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

\int_{- 1}^{1} 2 d x = 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 x} (\frac{2 x - y}{x + 2 y})

\frac{\partial ^{2}}{\partial x ^{2}} (x^{2} - y^{2})

\frac{\partial}{\partial x} ((4 - x^{2} - y^{2})^{\frac{1}{2}})

\frac{d}{d t} (\frac{1}{t ^{2} - 2})

\frac{\partial}{\partial x} (ln (4 + 3 x^{2} + 2 y^{2}))

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