What is area y=e^x,y=e^{-x},0,4 ?

The answer to area y=e^x,y=e^{-x},0,4 is (e^8+1)/(e^4)-2

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area y=e^x,y=e^{-x},0,4

area

y = e^{x}, y = e^{- x}, 0, 4

\frac{e ^{8} + 1}{e ^{4}} - 2

Decimal

52.61646 \dots

Solution steps

Apply the area formula:

\int_{0}^{4} e^{x} - e^{- x} d x

Solve

\int_{0}^{4} e^{x} - e^{- x} d x : \frac{e ^{8} + 1}{e ^{4}} - 2

The area is:

= \frac{e ^{8} + 1}{e ^{4}} - 2

\frac{x - 1}{2 - x}

x \to 1 + lim (5 - x^{2})

\frac{d}{d x} (ln (\frac{e ^{x} + e ^{- x}}{2}))

y y^{^{'}} = - 2 x (1 + y^{2}), y (0) = 1

\int 1. 8^{0.045 \cdot x} - 1 d x

What is area y=e^x,y=e^{-x},0,4 ?
The answer to area y=e^x,y=e^{-x},0,4 is (e^8+1)/(e^4)-2