What is area y=e^x,y=e^{-2x},x=ln(3) ?

The answer to area y=e^x,y=e^{-2x},x=ln(3) is 14/9

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area y=e^x,y=e^{-2x},x=ln(3)

area

y = e^{x}, y = e^{- 2 x}, x = ln (3)

\frac{14}{9}

Decimal Notation

1.55555 \dots

Solution steps

Apply the area formula:

\int_{0}^{l n (3)} e^{x} - e^{- 2 x} d x

Solve

\int_{0}^{l n (3)} e^{x} - e^{- 2 x} d x : \frac{14}{9}

The area is:

= \frac{14}{9}

\frac{d}{d x} ((cos (x))^{2} - 2 sin (x))

y = \frac{6}{sin ( x ) + cos ( x )}, (0, 6)

x \to 1 + lim (\frac{4}{2 x - 2})

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

x \to \infty lim (\frac{ln ( x ^{5} )}{13 x})

What is area y=e^x,y=e^{-2x},x=ln(3) ?
The answer to area y=e^x,y=e^{-2x},x=ln(3) is 14/9