I hope u can see it but I need help plz

For <img src="https://tex.z-dn.net/?f=e%5E%7B-x%5E2%2F2%7D" id="TexFormula1" title="e^{-x^2/2}" alt="e^{-x^2/2}" align="absmiddl

nevsk [136]

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.

3 0

2 years ago

This is 12 points!!!! please me help i don't undestand it

Paha777 [63]

Answer:

10,36,3,144,6 and 24 in that order

Step-by-step explanation:

sorry if its wrong

3 0

1 year ago

What is 3 2/5 + 2 1/10?

sdas [7]

Answer $\frac{11}{2}$ 

Step-by-step explanation:

7 0

3 years ago

HELPPPPPPPPPPPPPPPPPPPPPPPPPPPP! Is 16x^2 - 56x + 49 a perfect square trinomial? How do you know? If it is, factor the expressio

jasenka [17]

No need to fear, thehotdogman93 is here!

The first step is to get rid of those very large numbers. It's going to be very difficult to factor unless we can bring those high numbers down. So lets see if we can factor each term.

So after dividing 49 with every single digit. The only number that divides evenly is 7 and one, and 16 isnt divisible evenly by 7 so that didn't work. Looks like we're gonna have to work with these big numbers.

There is something interesting though about these numbers. 16 and 49 are both perfect squares. 16 is the same as 4^2 and 49 is the same as 7^2. So we can factor the whole trinomial as:

${(4x - 7)}^{2}$

If we were to expand this out as:

$(4x - 7)(4x - 7)$

and multiply it back into the original form. It would match with the expression we started with. The 4's would multiply back into 16x^2 and the 7's would multiply back into 49.

Additionally 4 * -7 is -28, so you can combine two -28x's into the -56x term in the original trinomial.

Thus, the answer is yes you can, and the answer is:

${(4x - 7)}^{2}$

6 0

2 years ago

A team played 20 games and won 18. what % did they win?

maxonik [38]

Answer: 90%

Reasoning: To solve this divide 18 by 20

18/20= 0.9

Now multiply 20 by 0.9 to check

20 x 0.9= 18

8 0

2 years ago