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

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

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

area

y = e^{x}, y = e^{- x}, x = - 2, x = 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

\int \frac{3 π}{2} cos (\frac{3 π x}{2}) d x

\frac{d}{d x} (arcsec (3 x + 2))

x \to \frac{7 π}{2} - lim (sec (x))

f (x) = \frac{2 x}{3 - 4 x ^{3}}, at x = \frac{1}{2}

\frac{\partial}{\partial x} (\frac{2}{1 - 3 x ^{3}})

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