What is area f(x)=1-e^{-x/3},2,3 ?

The answer to area f(x)=1-e^{-x/3},2,3 is 1-(3(e-e^{2/3}))/(e^{5/3)}

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area f(x)=1-e^{-x/3},2,3

area

f (x) = 1 - e^{- \frac{x}{3}}, 2, 3

1 - \frac{3 ( e - e ^{\frac{2}{3}} )}{e ^{\frac{5}{3}}}

Decimal

0.56338 \dots

Solution steps

Apply the area formula:

\int_{2}^{3} 1 - e^{- \frac{x}{3}} d x

Solve

\int_{2}^{3} 1 - e^{- \frac{x}{3}} d x : 1 - \frac{3 ( e - e ^{\frac{2}{3}} )}{e ^{\frac{5}{3}}}

The area is:

= 1 - \frac{3 ( e - e ^{\frac{2}{3}} )}{e ^{\frac{5}{3}}}

\int \frac{x ^{2} + x + 1}{x} d x

f (x) = ln (9 x)

f (x) = - 4 x^{2} + 6, at x = - 1

\frac{d y}{d x} = - \frac{( x e ^{- x^{2}} )}{y}

\int x^{(\frac{3}{2})} d x

What is area f(x)=1-e^{-x/3},2,3 ?
The answer to area f(x)=1-e^{-x/3},2,3 is 1-(3(e-e^{2/3}))/(e^{5/3)}