What is area 0.5(e^x+e^{-x}),0<= x<= ln(2) ?

The answer to area 0.5(e^x+e^{-x}),0<= x<= ln(2) is 0.74999…

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area 0.5(e^x+e^{-x}),0<= x<= ln(2)

area

0.5 (e^{x} + e^{- x}), 0 \leq x \leq ln (2)

0.74999 \dots

Solution steps

Apply the area formula:

\int_{0}^{l n (2)} 0.5 e^{x} + e^{- x} d x

Solve

\int_{0}^{l n (2)} 0.5 e^{x} + e^{- x} d x : 0.74999 \dots

The area is:

= 0.74999 \dots

\frac{d}{d x} (\frac{lo g _{10} ( x )}{ln ( x )} + 2 x)

y^{^{''}} + 5 y^{^{'}} + 6 y = 2 e^{x}

\frac{d}{d x} (in (sec (x)))

n = 1 \sum \infty \frac{2}{e ^{n - 3}}

y^{^{''}} - \frac{2}{x} y^{^{'}} - \frac{4}{x ^{2}} y = 3 x^{2}

What is area 0.5(e^x+e^{-x}),0<= x<= ln(2) ?
The answer to area 0.5(e^x+e^{-x}),0<= x<= ln(2) is 0.74999…