Solve the system of equations by graphing. {y=x+1 y=2x

Mathematics

1 answer:

lawyer [7]3 years ago

8 0

Answer:

x=1, y=2

Step-by-step explanation:

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Find the indefinite integral using the substitution provided.

Nady [450]

Answer: $7\text{Ln}\left(e^{2x}+10\right)+C$

This is the same as writing 7*Ln( e^(2x) + 10) + C

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Explanation:

Start with the equation $u = e^{2x}+10$

Apply the derivative and multiply both sides by 7 like so

$u = e^{2x}+10\\\\\frac{du}{dx} = 2e^{2x}\\\\7\frac{du}{dx} = 7*2e^{2x}\\\\7\frac{du}{dx} = 14e^{2x}\\\\7du = 14e^{2x}dx\\\\$

The "multiply both sides by 7" operation was done to turn the 2e^(2x) into 14e^(2x)

This way we can do the following substitutions:

$\displaystyle \int \frac{14e^{2x}}{e^{2x}+10}dx\\\\\\\displaystyle \int \frac{1}{e^{2x}+10}14e^{2x}dx\\\\\\\displaystyle \int \frac{1}{u}7du\\\\\\\displaystyle 7\int \frac{1}{u}du\\\\\\$

Integrating leads to

$\displaystyle 7\int \frac{1}{u}du\\\\\\7\text{Ln}\left(u\right)+C\\\\\\7\text{Ln}\left(e^{2x}+10\right)+C\\\\\\$

Be sure to replace 'u' with e^(2x)+10 since it's likely your teacher wants a function in terms of x. Also, do not forget to have the plus C at the end. This is a common mistake many students forget to do.

To verify the answer, you can apply the derivative to it and you should get back to the original integrand of $\frac{14e^{2x}}{e^{2x}+10}$