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}

Decimal Notation

3.68373 \dots

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}

2 y^{^{'}} + 9 y = 3

\int_{1}^{2} 7 \frac{ln ( x )}{x ^{2}} d x

\int (e^{x^{2}})^{2} d x

\frac{d}{d x} ((3 x + 5)^{3})

x \to 4 lim (3 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