What is area y=7e^x,y=7e^{-x},x=1 ?

The answer to area y=7e^x,y=7e^{-x},x=1 is 7e+7/e-14

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area y=7e^x,y=7e^{-x},x=1

area

y = 7 e^{x}, y = 7 e^{- x}, x = 1

7 e + \frac{7}{e} - 14

Solution steps

Apply the area formula:

\int_{0}^{1} 7 e^{x} - 7 e^{- x} d x

Solve

\int_{0}^{1} 7 e^{x} - 7 e^{- x} d x : 7 e + \frac{7}{e} - 14

The area is:

= 7 e + \frac{7}{e} - 14

y^{^{'}} = e^{4 x + 3 y}

n = 0 \sum \infty 3^{n}

\frac{\partial}{\partial y} (x^{2} ln (y + 1))

(- 1, - 2), (0, 1)

f (t) = 4 t^{3} cos (t)

What is area y=7e^x,y=7e^{-x},x=1 ?
The answer to area y=7e^x,y=7e^{-x},x=1 is 7e+7/e-14