NSB Integration Q (1 Viewer)

hmim · Jun 16, 2023

How is the solution to this question found, particularly this part?

011235 · Jun 16, 2023

split the integral into a to 0 and 0 to -a. then sub x=-u in the second integral, take the negative out the front and flip the bounds and dummy variable for x=u

lmk if you need full working

Drongoski · Jun 16, 2023

Or (not shorter):

$\int \frac{e^x}{e^x + e^{-x}} dx = \int \frac {e^x \cdot e^x}{e^{2x} + 1} dx = \frac {1}{2} \int \frac {1}{e^{2x} + 1} d(e^{2x} + 1) = \frac {1}{2} ln(e^{2x} + 1) + C \\ \\ \therefore \int ^a _{-a} \frac {e^x}{e^x + e^{-x}}dx = \frac {1}{2}\left [ ln(e^{2x} + 1)\right ]^a _{-a} = \frac {1}{2}[ln \frac {e^{2a} + 1}{e^{-2a} + 1}]\\ \\ = \frac {1}{2} ln[\frac {(e^{2a}+1)\cdot e^{2a}}{(e^{2a}+1)}] = \frac {1}{2}ln e^{2a} = a$

hmim · Jun 16, 2023

011235 said:
split the integral into a to 0 and 0 to -a. then sub x=-u in the second integral, take the negative out the front and flip the bounds and dummy variable for x=u

lmk if you need full working

hmm can you show full working please?

hmim · Jun 16, 2023

Drongoski said:
Or (not shorter):

$\int \frac{e^x}{e^x + e^{-x}} dx = \int \frac {e^x \cdot e^x}{e^{2x} + 1} dx = \frac {1}{2} \int \frac {1}{e^{2x} + 1} d(e^{2x} + 1) = \frac {1}{2} ln(e^{2x} + 1) + C \\ \\ \therefore \int ^a _{-a} \frac {e^x}{e^x + e^{-x}}dx = \frac {1}{2}\left [ ln(e^{2x} + 1)\right ]^a _{-a} = \frac {1}{2}[ln \frac {e^{2a} + 1}{e^{-2a} + 1}]\\ \\ = \frac {1}{2} ln[\frac {(e^{2a}+1)\cdot e^{2a}}{(e^{2a}+1)}] = \frac {1}{2}ln e^{2a} = a$

Thank you very much, that makes sense!

011235 · Jun 17, 2023

$\int ^a _{-a} \frac{e^x}{e^x + e^{-x}} dx = \int ^a _0 \frac{e^x}{e^x + e^{-x}} dx + \int ^0 _{-a} \frac{e^x}{e^x + e^{-x}} dx$

Sub x = -u (dx = -du) in the second integral
$=\int^a_0 \frac{e^x}{e^x + e^{-x}} dx + \int ^0 _a \frac{e^{-u}}{e^{-u} + e^u} \times -du = \int^a_0 \frac{e^x}{e^x + e^{-x}} dx - \int^0 _a \frac{e^{-u}}{e^{-u} + e^u} du = \int^a_0 \frac{e^x}{e^x + e^{-x}} dx + \int^a_0 \frac{e^{-u}}{e^{-u} + e^u}du = \int^a_0 \frac{e^x}{e^x + e^{-x}} dx + \int^a_0 \frac{e^{-x}}{e^{-x} + e^x}dx$

The last step is just changing the variable (this is possible for all definite integrals because the name of the variable doesnt matter)

chilli 412 · Jun 17, 2023

011235 said:
$\int ^a _{-a} \frac{e^x}{e^x + e^{-x}} dx = \int ^a _0 \frac{e^x}{e^x + e^{-x}} dx + \int ^0 _{-a} \frac{e^x}{e^x + e^{-x}} dx$

Sub x = -u (dx = -du) in the second integral
$=\int^a_0 \frac{e^x}{e^x + e^{-x}} dx + \int ^0 _a \frac{e^{-u}}{e^{-u} + e^u} \times -du = \int^a_0 \frac{e^x}{e^x + e^{-x}} dx - \int^0 _a \frac{e^{-u}}{e^{-u} + e^u} du = \int^a_0 \frac{e^x}{e^x + e^{-x}} dx + \int^a_0 \frac{e^{-u}}{e^{-u} + e^u}du = \int^a_0 \frac{e^x}{e^x + e^{-x}} dx + \int^a_0 \frac{e^{-x}}{e^{-x} + e^x}dx$

The last step is just changing the variable (this is possible for all definite integrals because the name of the variable doesnt matter)

that switching from u to x with definite integrals always gives me the ick... feels illegal yet works

NSB Integration Q (1 Viewer)

hmim

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011235

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Drongoski

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hmim

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hmim

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011235

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chilli 412

oo la la

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