What is area y=e^{-2x}+1,y=0,x=0,x=3 ?

The answer to area y=e^{-2x}+1,y=0,x=0,x=3 is (e^6-1)/(2e^6)+3

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area y=e^{-2x}+1,y=0,x=0,x=3

area

y = e^{- 2 x} + 1, y = 0, x = 0, x = 3

\frac{e ^{6} - 1}{2 e ^{6}} + 3

Decimal Notation

3.49876 \dots

Solution steps

Apply the area formula:

\int_{0}^{3} e^{- 2 x} + 1 d x

Solve

\int_{0}^{3} e^{- 2 x} + 1 d x : \frac{e ^{6} - 1}{2 e ^{6}} + 3

The area is:

= \frac{e ^{6} - 1}{2 e ^{6}} + 3

\frac{\partial}{\partial t} (3 e^{s i n (x + 4 c t)})

f (x) = x^{6 x}

y^{^{'}} = \frac{( 4 x ^{3} + y ^{3} )}{x y ^{2}}

n = 0 \sum \infty \frac{1}{2 n !}

y^{^{'}} = cos^{2} (y) ln (x)

What is area y=e^{-2x}+1,y=0,x=0,x=3 ?
The answer to area y=e^{-2x}+1,y=0,x=0,x=3 is (e^6-1)/(2e^6)+3