What is area y=e^{-x}sin(x),y=-e^{-x}sin(x),0<= x<= pi ?

The answer to area y=e^{-x}sin(x),y=-e^{-x}sin(x),0<= x<= pi is 1/(e^pi)+1

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area y=e^{-x}sin(x),y=-e^{-x}sin(x),0<= x<= pi

area

y = e^{- x} sin (x), y = - e^{- x} sin (x), 0 \leq x \leq π

\frac{1}{e ^{π}} + 1

Decimal Notation

1.04321 \dots

Solution steps

Apply the area formula:

\int_{0}^{π} 2 e^{- x} ∣ sin (x) ∣ d x

Solve

\int_{0}^{π} 2 e^{- x} ∣ sin (x) ∣ d x : \frac{1}{e ^{π}} + 1

The area is:

= \frac{1}{e ^{π}} + 1

\int \frac{cos ( 7 x )}{1 + sin ^{2} ( 7 x )} d x

f (x) = x^{2} (x - 2)^{4}

\frac{d y}{d x} = \frac{y}{x ^{2} + 1}

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

\frac{d}{d x} (\frac{- 2 x - y}{x + 2 y})

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