What is area y=e^x,y=e^{4x},x=1 ?

The answer to area y=e^x,y=e^{4x},x=1 is -e+1+(e^4-1)/4

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area y=e^x,y=e^{4x},x=1

area

y = e^{x}, y = e^{4 x}, x = 1

- e + 1 + \frac{e ^{4} - 1}{4}

Decimal

11.68125 \dots

Solution steps

Apply the area formula:

\int_{0}^{1} e^{x} - e^{4 x} d x

Solve

\int_{0}^{1} e^{x} - e^{4 x} d x : - e + 1 + \frac{e ^{4} - 1}{4}

The area is:

= - e + 1 + \frac{e ^{4} - 1}{4}

\frac{1}{( ( s - 1 ) ^{2} + 4 )}

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

y^{^{''}} - 4 y + 4 y = 0

y^{^{'}} = x (2 - y)

x \to - \infty lim (- 3 e^{x})

What is area y=e^x,y=e^{4x},x=1 ?
The answer to area y=e^x,y=e^{4x},x=1 is -e+1+(e^4-1)/4