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

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

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

area

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

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

Decimal Notation

134.14293 \dots

Solution steps

Apply the area formula:

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

Solve

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

The area is:

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

\frac{d}{d x} (- x cos (x) - x)

\frac{d}{d x} ((x^{3} + 7 x - 1) (5 x + 2))

x \to 1 lim (\frac{0}{0})

f (x) = x^{2} - 4 x + 3

\frac{\partial}{\partial x} (x^{4} + x^{2} y^{2} + y^{5} + x + y)

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