Question

Sketch the region enclosed by the curves and find its area. \[ y=e^{x}, y=e^{2 x}, x=0, x=\ln 3 \] Area \( = \) You have attempted this problem 0 times. You have 5 attempts remaining.

Answer 1

Area
`\( = (1)/(3)(e^(\ln 3)-1)(e^(\ln 3)-e^0) = (2)/(3)(3-(1)/(e)) \)` This integral captures the area between the exponential curves within the specified bounds along the x-axis, resulting in the final area of
`\((2)/(3)(3-(1)/(e))\).`

Step-by-step explanation:

To find the enclosed region's area between the curves
`\(y = e^x\) and \(y = e^(2x)\) bounded by \(x = 0\) and \(x = \ln 3\),`we utilize definite integrals along the x-axis. First, identify the intersection points of the curves by setting
`\(e^x = e^(2x)\), yielding \(x = \ln 3\).`

Integrating
`\(e^(2x)\) and \(e^x\)`with respect to
`\(x\) from \(0\) to \(\ln 3\)`, we form the definite integral
`\(\int_(0)^(\ln 3) (e^(2x) - e^x) \,dx\)` to calculate the enclosed area. Simplifying the integral gives us
`\((1)/(3)(e^(3x)-e^(2x))\)` evaluated from (0) to (ln 3).

Substituting the bounds into the antiderivative and simplifying, we obtain
`\((1)/(3)(e^(\ln 3)-1)(e^(\ln 3)-e^0)\).` Further simplification yields
`\((1)/(3)(3-1)(3-(1)/(e)) = (2)/(3)(3-(1)/(e))\),`representing
`\((1)/(3)(e^(\ln 3)-1)(e^(\ln 3)-e^0)\)`the area between the curves.

This integral captures the area between the exponential curves within the specified bounds along the x-axis, resulting in the final area of
`\((2)/(3)(3-(1)/(e))\).`

"".

Answer 2

The region enclosed by the curves
`\(y = e^x\), \(y = e^(2x)\), \(x = 0\), and \(x = \ln 3\),` is
`\(-(5)/(2)\).`

The integral for the area between the curves.

First, let's find the points of intersection of the curves

`\(y = e^x\) and \(y = e^(2x)\):`

`\[ e^x = e^(2x) \]`

Solving for x, we get
`\(x = 0\) and \(x = \ln 3\)`.

Now, the integral for the area between the curves is given by:

`\[ \text{Area} = \int_(a)^(b) \left|f(x) - g(x)\right| \,dx \]`

where a and b are the x-coordinates of the points of intersection.

In this case, the area is:

`\[ \text{Area} = \int_(0)^(\ln 3) \left|e^(2x) - e^x\right| \,dx \]`

Now, we can simplify this expression:

`\[ \text{Area} = \int_(0)^(\ln 3) (e^(2x) - e^x) \,dx \]`

Now, integrate with respect to x:

`\[ \text{Area} = (1)/(2)e^(2x) - e^x \Big|_(0)^(\ln 3) \]`

Evaluate the expression at
`\(\ln 3` and subtract the value at 0:

`\[ \text{Area} = (1)/(2)e^(2(\ln 3)) - e^(\ln 3) - \left((1)/(2)e^0 - e^0\right) \]`

`\[ \text{Area} = (1)/(2)(3^2) - 3 - \left((1)/(2) - 1\right) \]`

`\[ \text{Area} = (9)/(2) - 3 + (1)/(2) \]`

`\[ \text{Area} = -(5)/(2) \]`

So, the area of the region enclosed by the curves is
`\(-(5)/(2)\).` Note that the negative sign indicates that the area is measured below the x-axis.