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

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

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

area

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

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

Decimal Notation

1.13533 \dots

Solution steps

Apply the area formula:

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

Solve

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

The area is:

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

\frac{d}{d x} (arccos (\frac{x}{π}))

\int sec^{2} (x) - 6 d x

\frac{\partial}{\partial y} (3 x^{2} y^{4} + 1)

\frac{\partial}{\partial x} (4 x^{3} y^{2} - 4 x^{2} + y^{6} + 1)

\int_{0}^{1} x cos (x) d x

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