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

The integral from-1 to 1 of (e^x-(x^2-1)) is e-1/e+4/3

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

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

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

Solution steps

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

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

\int_{- 1}^{1} x^{2} - 1 d x = - \frac{4}{3}

= e - \frac{1}{e} - (- \frac{4}{3})

Simplify

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

\int \frac{3 x + 1}{x ^{2} + 1} d x

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

x \to 1 lim (sin (\frac{π x}{2}))

\int - 3 \cdot x + 2 x d x

\int \frac{x ^{2}}{x ^{2} - 4} d x

What is the integral from-1 to 1 of (e^x-(x^2-1)) ?
The integral from-1 to 1 of (e^x-(x^2-1)) is e-1/e+4/3