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

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

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

area

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

e - 1 - \frac{e ^{2} - 1}{2 e}

Decimal

0.54308 \dots

Solution steps

Apply the area formula:

\int_{0}^{1} e^{x} - e^{2 x - 1} d x

Solve

\int_{0}^{1} e^{x} - e^{2 x - 1} d x : e - 1 - \frac{e ^{2} - 1}{2 e}

The area is:

= e - 1 - \frac{e ^{2} - 1}{2 e}

\int cos (\frac{1}{2}) d x

\int (\frac{3}{2 x - 1}) d x

\frac{d y}{d x}, x^{2} + y^{2} = 2 x y

\frac{d}{d t} (2 - t)

y^{^{''}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = 1

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