What is area y=e^x,y=e^{-x},[-2,1] ?

The answer to area y=e^x,y=e^{-x},[-2,1] is e^2+(1+e^3+e)/(e^2)-4

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area y=e^x,y=e^{-x},[-2,1]

area

y = e^{x}, y = e^{- x}, [- 2, 1]

e^{2} + \frac{1 + e ^{3} + e}{e ^{2}} - 4

Decimal Notation

6.61055 \dots

Solution steps

Apply the area formula:

\int_{- 2}^{1} e^{x} - e^{- x} d x

Solve

\int_{- 2}^{1} e^{x} - e^{- x} d x : e^{2} + \frac{1 + e ^{3} + e}{e ^{2}} - 4

The area is:

= e^{2} + \frac{1 + e ^{3} + e}{e ^{2}} - 4

\frac{d}{d x} (1 + 2 sin (x))

\frac{d}{d x} (\frac{( x ^{3} )}{3} + \frac{1}{4 x})

\int \frac{1}{x ( x ^{999} + 1 ) ^{2}} d x

f (x) = 2 x^{3}, (1, 2)

\frac{d}{d y} (ln (\frac{( 2 y + 1 ) ^{4}}{y ^{2} + 1}))

What is area y=e^x,y=e^{-x},[-2,1] ?
The answer to area y=e^x,y=e^{-x},[-2,1] is e^2+(1+e^3+e)/(e^2)-4