What is area y=e^{3x},y=e^x,[0,1] ?

The answer to area y=e^{3x},y=e^x,[0,1] is (e^3-1)/3-e+1

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area y=e^{3x},y=e^x,[0,1]

area

y = e^{3 x}, y = e^{x}, [0, 1]

\frac{e ^{3} - 1}{3} - e + 1

Decimal Notation

4.64356 \dots

Solution steps

Apply the area formula:

\int_{0}^{1} e^{3 x} - e^{x} d x

Solve

\int_{0}^{1} e^{3 x} - e^{x} d x : \frac{e ^{3} - 1}{3} - e + 1

The area is:

= \frac{e ^{3} - 1}{3} - e + 1

x y^{^{'}} - y = y^{3}

\int_{0}^{9} \frac{x}{( x + 9 ) ^{4}} d x

\int_{1}^{2} (x - 2)^{- 2} d x

\frac{2}{x ^{6}}

\int_{0}^{1} sin (π x cos (π) x) d x

What is area y=e^{3x},y=e^x,[0,1] ?
The answer to area y=e^{3x},y=e^x,[0,1] is (e^3-1)/3-e+1