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

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

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

area

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

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

Decimal

2.76219 \dots

Solution steps

Apply the area formula:

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

Solve

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

The area is:

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

y = \frac{f ( x )}{g ( x )}

\frac{\partial}{\partial y} (x^{2} y + 4 y - 7)

f (x) = (2 x - 3)^{l n (x)}

\int (x^{1.1} + 7 x^{2.5}) d x

x \to 2 lim (3 x^{3} - 1)

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