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3 years ago

I NEEDS THE HELP RIGHT NOW

Mathematics

1 answer:

leva [86]3 years ago

3 0

Answer:idlk

Step-by-step explanation:

idlk

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Determine whether the integral converges.

Kryger [21]

You have one mistake which occurs when you integrate $\dfrac1{1-p^2}$ . The antiderivative of this is not in terms of $\tan^{-1}p$ . Instead, letting $p=\sin r$ (or $\cos r$ , if you want to bother with more signs) gives $\mathrm dp=\cos r\,\mathrm dr$ , making the indefinite integral equality

$\displaystyle-\frac12\int\frac{\mathrm dp}{1-p^2}=-\frac12\int\frac{\cos r}{1-\sin^2r}\,\mathrm dr=-\frac12\int\sec r\,\mathrm dr=\ln|\sec r+\tan r|+C$

and then compute the definite integral from there.

$-\dfrac12\ln|\sec r+\tan r|\stackrel{r=\sin^{-1}p}=-\dfrac12\ln\left|\dfrac{1+p}{\sqrt{1-p^2}}=\ln\left|\sqrt{\dfrac{1+p}{1-p}}\right|$
$\stackrel{p=u/2}=-\dfrac12\ln\left|\sqrt{\dfrac{1+\frac u2}{1-\frac u2}}\right|=-\dfrac12\ln\left|\sqrt{\dfrac{2+u}{2-u}}\right|$
$\stackrel{u=x+1}=-\dfrac12\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|$
$\implies-\dfrac12\displaystyle\lim_{t\to\infty}\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|\bigg|_{x=2}^{x=t}=-\frac12\left(\ln|-1|-\ln\left|\sqrt{\frac5{-1}}\right|\right)=\dfrac{\ln\sqrt5}2=\dfrac{\ln5}4$

Or, starting from the beginning, you could also have found it slightly more convenient to combine the substitutions in one fell swoop by letting $x+1=2\sec y$ . Then $\mathrm dx=2\sec y\tan y\,\mathrm dy$ , and the integral becomes

$\displaystyle\int_2^\infty\frac{\mathrm dx}{(x+1)^2-4}=\int_{\sec^{-1}(3/2)}^{\pi/2}\frac{2\sec y\tan y}{4\sec^2y-4}\,\mathrm dy$
$\displaystyle=\frac12\int_{\sec^{-1}(3/2)}^{\pi/2}\csc y\,\mathrm dy$
$\displaystyle=-\frac12\ln|\csc y+\cot y|\bigg|_{y=\sec^{-1}(3/2}}^{y=\pi/2}$
$\displaystyle=-\frac12\lim_{t\to\pi/2^-}\ln|\csc y+\cot y|\bigg|_{y=\sec^{-1}(3/2)}^{y=t}$
$\displaystyle=-\frac12\left(\lim_{t\to\pi/2^-}\ln|\csc t+\cot t|-\ln\frac5{\sqrt5}\right)$
$=\dfrac{\ln\sqrt5}2-\dfrac{\ln|1|}2$
$=\dfrac{\ln5}4$

Another way to do this is to notice that the integrand's denominator can be factorized.

$x^2+2x-3=(x+3)(x-1)$

So,

$\dfrac1{x^2+2x-3}=\dfrac1{(x+3)(x-1)}=\dfrac14\left(\dfrac1{x-1}-\dfrac1{x+3}\right)$

There are no discontinuities to worry about since you're integrate over $[2,\infty)$ , so you can proceed with integrating straightaway.

$\displaystyle\int_2^\infty\frac{\mathrm dx}{x^2+2x-3}=\frac14\lim_{t\to\infty}\int_2^t\left(\frac1{x-1}-\frac1{x+3}\right)\,\mathrm dx$
$=\displaystyle\frac14\lim_{t\to\infty}(\ln|x-1|-\ln|x+3|)\bigg|_{x=2}^{x=t}$
$=\displaystyle\frac14\lim_{t\to\infty}\ln\left|\frac{x-1}{x+3}\right|\bigg|_{x=2}^{x=t}$
$=\displaystyle\frac14\left(\lim_{t\to\infty}\ln\left|\frac{t-1}{t+3}\right|-\ln\frac15\right)$
$=\displaystyle\frac14\left(\ln1-\ln\frac15\right)$
$=-\dfrac14\ln\dfrac15=\dfrac{\ln5}4$

Just goes to show there's often more than one way to skin a cat...

7 0

2 years ago

Each time a component is tested, the trial is a success (S) or failure (F). Suppose the component is tested repeatedly until a s

ehidna [41]

Answer:

Step-by-step explanation: Let S represent the sample space , then the possible trials are: S= {SSS, FSSS, FFSSS, FFFSSS, FFFFSSS, …}

Y = the number of trials to be tested.

The possible values of Y are P ={x ∈N , x≥3}

Therefore the five smallest possible values of Y are :

1. Y{SSS} ........ 3 outcomes

2. Y{FSSS} ...... 4 outcomes

3. Y{FFSSS}....... 5 outcomes

4. Y{FFFSSS} ...... 6 outcomes

5. Y{FFFFSSS} ..... 7 outcomes

3 0

3 years ago

Which statements about the function in item 3 are true? select all that apply

Aleks04 [339]

Answer:

C. Some input values are integers

Step-by-step explanation:

4 0

3 years ago

-8m<-72 need help finding m!! Plz

vladimir2022 [97]

Answer:

m > 9

Step-by-step explanation:

-8m<-72

Divide each side by -8, remembering to flip the inequality

-8m/-8>-72/-8

m > 9

4 0

3 years ago

Read 2 more answers

According to the College Board, SAT critical reading scores from the 2015 school year for high school students in the United Sta

vagabundo [1.1K]

Answer:

61.1% of high school student who took the SAT in 2015 will have a critical reading SAT score between 400 and 600 points.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 495

Standard Deviation, σ = 116

We are given that the distribution of scores from the 2015 school year for high school students in the United States is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(SAT score between 400 and 600 points)

$P(400 \leq x \leq 600) = P(\displaystyle\frac{400 - 495}{116} \leq z \leq \displaystyle\frac{600-495}{116}) = P(-0.818 \leq z \leq 0.905)\\\\= P(z \leq 0.905) - P(z < -0.818)\\= 0.8173 - 0.2064= 0.6109 \approx 61.1\%$

$P(400 \leq x \leq 600) = 61.1\%$

Thus, 61.1% of high school student who took the SAT in 2015 will have a critical reading SAT score between 400 and 600 points.

3 0

3 years ago