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Area between the curve y=5e^(3x) and y=e^(5x) + 2 from x = 0 to x =1

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To find the area between the curves \(y = 5e^{3x}\) and \(y = e^{5x} + 2\) from \(x = 0\) to \(x = 1\), you need to set up an integral to compute the difference between the two curves over the given interval.

The area between two curves \(y = f(x)\) and \(y = g(x)\) over the interval \([a, b]\) is given by the definite integral:

\[A = \int_{a}^{b} |f(x) - g(x)| \, dx.\]

In this case, you want to find the area between the curves \(y = 5e^{3x}\) and \(y = e^{5x} + 2\) over the interval \([0, 1]\). Thus, the area can be calculated as follows:

\[A = \int_{0}^{1} |(5e^{3x}) - (e^{5x} + 2)| \, dx.\]

First, let's simplify the expression inside the absolute value:

\[|5e^{3x} - (e^{5x} + 2)| = |5e^{3x} - e^{5x} - 2|.\]

Now, you need to integrate this expression over the interval \([0, 1]\):

\[A = \int_{0}^{1} |5e^{3x} - e^{5x} - 2| \, dx.\]

This integral involves a piecewise function, because the sign of \(5e^{3x} - e^{5x} - 2\) changes at some point within the interval \([0, 1]\). To handle this, you need to determine the point where the sign changes (i.e., where \(5e^{3x} - e^{5x} - 2 = 0\)), and then break the integral into two parts accordingly.

Solve for \(x\) when \(5e^{3x} - e^{5x} - 2 = 0\):

\[5e^{3x} = e^{5x} + 2.\]

\[e^{3x}(5 - e^{2x}) = 2.\]

Now, you would need to solve this transcendental equation for \(x\), which might involve numerical methods.

Once you find the value of \(x\) where the sign changes, let's call it \(c\), you can split the integral into two parts:

\[A = \int_{0}^{c} (e^{5x} + 2 - 5e^{3x}) \, dx + \int_{c}^{1} (5e^{3x} - e^{5x} - 2) \, dx.\]

Now, you can integrate each part separately and then sum up the results to find the total area between the curves.

Please note that solving the equation \(5e^{3x} - e^{5x} - 2 = 0\) for \(x\) analytically might not be straightforward, and you might need to use numerical methods to approximate the value of \(c\). Once you have the value of \(c\), you can proceed with the integration and calculations.

User Chris Barker
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