What is the integral from-1 to 1 of e^y-(y^2-9) ?

The integral from-1 to 1 of e^y-(y^2-9) is e-1/e+52/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-1 to 1 of e^y-(y^2-9)

\int_{- 1}^{1} e^{y} - (y^{2} - 9) d y

e - \frac{1}{e} + \frac{52}{3}

Decimal

19.68373 \dots

Solution steps

\int_{- 1}^{1} e^{y} - (y^{2} - 9) 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} - 9 d y

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

\int_{- 1}^{1} y^{2} - 9 d y = - \frac{52}{3}

= e - \frac{1}{e} - (- \frac{52}{3})

Simplify

= e - \frac{1}{e} + \frac{52}{3}

(14 - 6 y + e - 3 x) d x - 2 d y = 0

\int_{0}^{6} \frac{π}{8} (6 - x)^{2} d x

\int sin^{4} (x co) s^{2} x d x

y = \frac{cos ( x )}{x ^{7}}

x \to 0 lim (\frac{sin ( 2 x )}{x})

What is the integral from-1 to 1 of e^y-(y^2-9) ?
The integral from-1 to 1 of e^y-(y^2-9) is e-1/e+52/3