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

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

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

area

y = e^{2 x}, x = 0, x = 3

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

Decimal Notation

201.21439 \dots

Solution steps

Apply the area formula:

\int_{0}^{3} e^{2 x} d x

Solve

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

The area is:

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

\int \frac{e ^{6 t}}{t} d t

\frac{d}{d x} (e^{(\frac{- k}{x})})

2 (sin (x^{2}) + 2 x^{2} cos (x^{2}))

f (x) = \frac{1}{3 x}, at x = 9

x \to - 7 lim (\frac{7 - ∣ x ∣}{7 + x})

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