What is the integral from-1 to 1 of (y^2-7)-(e^y) ?

The integral from-1 to 1 of (y^2-7)-(e^y) is -e+1/e-40/3

Popular >

integral from-1 to 1 of (y^2-7)-(e^y)

\int_{- 1}^{1} (y^{2} - 7) - (e^{y}) d y

- e + \frac{1}{e} - \frac{40}{3}

Decimal Notation

- 15.68373 \dots

Solution steps

\int_{- 1}^{1} y^{2} - 7 - e^{y} 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} y^{2} d y - \int_{- 1}^{1} 7 d y - \int_{- 1}^{1} e^{y} d y

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

\int_{- 1}^{1} 7 d y = 14

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

= \frac{2}{3} - 14 - (e - \frac{1}{e})

Simplify

\frac{2}{3} - 14 - (e - \frac{1}{e}) : - e + \frac{1}{e} - \frac{40}{3}

= - e + \frac{1}{e} - \frac{40}{3}

\int_{0}^{1} \frac{1}{5 x - 10} d x

\int_{0}^{x} t^{3} d t

\int_{0}^{1} 6 (\frac{y}{x})^{4} d y

\int_{1}^{25} \frac{1}{1 + x} d x

\int_{1}^{\infty} \frac{1}{n ^{3}}

What is the integral from-1 to 1 of (y^2-7)-(e^y) ?
The integral from-1 to 1 of (y^2-7)-(e^y) is -e+1/e-40/3