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DaniilM [7]
3 years ago
5

For

e" class="latex-formula">, how would the substitution u=\frac{x}{\sqrt{2}}
Mathematics
1 answer:
nevsk [136]3 years ago
3 0
I'm assuming you're talking about the indefinite integral

\displaystyle\int e^{-x^2/2}\,\mathrm dx

and that your question is whether the substitution u=\dfrac x{\sqrt2} would work. Well, let's check it out:

u=\dfrac x{\sqrt2}\implies\mathrm du=\dfrac{\mathrm dx}{\sqrt2}
\implies\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt2\int e^{-(\sqrt2\,u)^2/2}\,\mathrm du
=\displaystyle\sqrt2\int e^{-u^2}\,\mathrm du

which essentially brings us to back to where we started. (The substitution only served to remove the scale factor in the exponent.)

What if we tried u=\sqrt t next? Then \mathrm du=\dfrac{\mathrm dt}{2\sqrt t}, giving

=\displaystyle\frac1{\sqrt2}\int \frac{e^{-(\sqrt t)^2}}{\sqrt t}\,\mathrm dt=\frac1{\sqrt2}\int\frac{e^{-t}}{\sqrt t}\,\mathrm dt

Next you may be tempted to try to integrate this by parts, but that will get you nowhere.

So how to deal with this integral? The answer lies in what's called the "error function" defined as

\mathrm{erf}(x)=\displaystyle\frac2{\sqrt\pi}\int_0^xe^{-t^2}\,\mathrm dt

By the fundamental theorem of calculus, taking the derivative of both sides yields

\dfrac{\mathrm d}{\mathrm dx}\mathrm{erf}(x)=\dfrac2{\sqrt\pi}e^{-x^2}

and so the antiderivative would be

\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt{\frac\pi2}\mathrm{erf}\left(\frac x{\sqrt2}\right)

The takeaway here is that a new function (i.e. not some combination of simpler functions like regular exponential, logarithmic, periodic, or polynomial functions) is needed to capture the antiderivative.
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2 years ago
A wallet contains $460 in $5, $10, and $20 bills. The number of $5 bills exceeds twice the number of $10 bills by 4, and the num
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Answer:

Number of $5 bills = 32

Number of $10 bills = 14

Number of $20 bills = 8

Step-by-step explanation:

Let x number of $5, y number of $10 and z number of $20

The number of $5 bills exceeds twice the number of $10 bills by 4.

Therefore, x = 2y + 4

The number of $20 bills is 6 fewer than the number of $10 bills.

Therefore, z = y - 6

A wallet contains $460 in $5, $10, and $20 bills.

Therefore,

5x + 10y + 20z = 460

Substitute x and y into equation

    5(2y+4) + 10y + 20(y-6) = 460

10y + 20 + 10y + 20y - 120 = 460

                           40y - 100 = 460

                                    40y = 460 + 100

                                    40y = 560

                                      y = 14

  • Put the value of y into x = 2y + 4 and solve for x

                                      x = 2(14) + 4

                                      x = 32

  • Put the value of y into z = y - 6 and solve for z

                                      z = 14 - 6

                                      z = 8

Hence, the each type of bills,

Number of $5 bills = 32

Number of $10 bills = 14

Number of $20 bills = 8

3 0
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Answer:

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