What is area y=e^x,y=e^{-2x},x=ln(4) ?

The answer to area y=e^x,y=e^{-2x},x=ln(4) is 81/32

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area y=e^x,y=e^{-2x},x=ln(4)

area

y = e^{x}, y = e^{- 2 x}, x = ln (4)

\frac{81}{32}

Decimal Notation

2.53125

Solution steps

Apply the area formula:

\int_{0}^{l n (4)} e^{x} - e^{- 2 x} d x

Solve

\int_{0}^{l n (4)} e^{x} - e^{- 2 x} d x : \frac{81}{32}

The area is:

= \frac{81}{32}

10 y^{^{''}} + 20 y^{^{'}} + 6260 y = 0

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

\frac{d}{d x} (\frac{5 x + 1}{4 x ^{2} + 1})

\frac{d}{d x} (lo g_{6} (5 x + 1))

\frac{d}{d x} ((arcsin (x)) d x)

What is area y=e^x,y=e^{-2x},x=ln(4) ?
The answer to area y=e^x,y=e^{-2x},x=ln(4) is 81/32