What is area y=e^x,y=e^{-x},x=1 ?

The answer to area y=e^x,y=e^{-x},x=1 is (e^2+1)/e-2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area y=e^x,y=e^{-x},x=1

area

y = e^{x}, y = e^{- x}, x = 1

\frac{e ^{2} + 1}{e} - 2

Decimal Notation

1.08616 \dots

Solution steps

Apply the area formula:

\int_{0}^{1} e^{x} - e^{- x} d x

Solve

\int_{0}^{1} e^{x} - e^{- x} d x : \frac{e ^{2} + 1}{e} - 2

The area is:

= \frac{e ^{2} + 1}{e} - 2

x e^{- 5 x}

\int \frac{x ^{3} + 3 x ^{2} + 6}{x ^{2} + 2} d x

\int \frac{x}{( x ^{2} + 4 )} d x

\int 4 + 3 x d x

\frac{\partial}{\partial x} (x + sin (y + z))

What is area y=e^x,y=e^{-x},x=1 ?
The answer to area y=e^x,y=e^{-x},x=1 is (e^2+1)/e-2