What is area y=e^x,y=2e^{-x}+1,x=0 ?

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

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area y=e^x,y=2e^{-x}+1,x=0

area

y = e^{x}, y = 2 e^{- x} + 1, x = 0

ln (2)

Decimal

0.69314 \dots

Solution steps

Apply the area formula:

\int_{0}^{l n (2)} e^{x} - (2 e^{- x} + 1) d x

Solve

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

The area is:

= ln (2)

\frac{\partial}{\partial x} (- 4 sin (2 x))

\frac{d y}{d x}, x^{3} + y^{3} = 36

(0, 1), (1, - 3)

x \to - \infty lim (x - 4)

\frac{2}{s ^{3} ( s ^{2} + 9 )}

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