What is the integral from-1 to 1 of e^y-(y^2-8) ?

The integral from-1 to 1 of e^y-(y^2-8) is e-1/e+46/3

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integral from-1 to 1 of e^y-(y^2-8)

\int_{- 1}^{1} e^{y} - (y^{2} - 8) d y

e - \frac{1}{e} + \frac{46}{3}

Decimal

17.68373 \dots

Solution steps

\int_{- 1}^{1} e^{y} - (y^{2} - 8) 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} - 8 d y

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

\int_{- 1}^{1} y^{2} - 8 d y = - \frac{46}{3}

= e - \frac{1}{e} - (- \frac{46}{3})

Simplify

= e - \frac{1}{e} + \frac{46}{3}

f (x) = \frac{1}{x}, at x = - 4

\int \frac{1}{2 x ^{2} ( x + 1 )} d x

x (1 - 9 x^{3})

x \to 2 + lim (2 x + 2 + x)

\int \frac{x ^{3}}{( x + 1 )} d x

What is the integral from-1 to 1 of e^y-(y^2-8) ?
The integral from-1 to 1 of e^y-(y^2-8) is e-1/e+46/3