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Liono4ka [1.6K]
3 years ago
8

Integrate 1/((x^5)*sqrt(9*x^2-1)) ...?

Mathematics
1 answer:
Korolek [52]3 years ago
5 0
Let 9x^2-1 = y^2 
<span>=> 18xdx = 2ydy </span>
<span>=> ydy = 9xdx </span>

<span>lower limit = sqrt(9*2/9 - 1) = sqrt(1) = 1 </span>
<span>upper limit = sqrt(9*4/9 - 1) = sqrt(3) </span>

<span>Int. [sqrt(2)/3,2/3] 1/(x^5(sqrt(9x^2-1)) dx </span>

<span>= Int. [sqrt(2)/3,2/3] xdx/(x^6(sqrt(9x^2-1)) </span>
<span>= 81* Int. [1,sqrt(3)] ydy/((y^2+1)^3y) </span>
<span>=81* Int. [1,sqrt(3)] dy/(y^2+1)^3 </span>

<span>y=tanz </span>
<span>dy = sec^2z dz </span>

<span>=81*Int [pi/4,pi/3] cos^4(z) dz </span>
<span>=81/4*int [pi/4,pi/3] (1+cos(2z))^2 dz </span>
<span>=81/4* Int. [pi/4,pi/3] (1+2cos(2z)+cos^2(2z)) dz </span>

<span>=81/4*(pi/3-pi/4) + 81/4*(sin(2pi/3)-sin(pi/2)) + 81/8 * (pi/3-pi/4) </span>
<span>+ 81/32 *(sin(-pi/3)-sin(pi)) </span>
<span>=81(pi/48+pi/96+1/4*(sqrt(3)/2 - 1) - 1/32 * sqrt(3)/2) </span>
<span>=81/32*(pi+3sqrt(3)-8)</span>
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Add a product row and a totals column:

\begin{array}{ | c | c | c | c | c | c | c | c | c | c |}\cline{1-10}x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 &\text{Total} \\ \cline{1-10} &&&&&&&&& \\ \text{P}(X=x) & \frac{1}{10} & \frac{1}{10} & \frac{1}{10} & \frac{3}{10} & \frac{1}{10} & \frac{1}{10} & \frac{1}{10} & \frac{1}{10} & 1\\ &&&&&&&&&\\ \cline{1-10} &&&&&&&&&\\\text{Product} & \frac{1}{10} & \frac{2}{10} & \frac{3}{10} & \frac{12}{10} & \frac{5}{10} & \frac{6}{10} & \frac{7}{10} & \frac{8}{10} & \frac{44}{10} \\ &&&&&&&&&\\\cline{1-10}\end{array}

(The Product is row is the product of the <u>score on the die</u> and <u>its probability</u>).

The expected value (EV) is the sum of the product of each outcome and its probability.  

Therefore, the expected value (EV) of this die is 44/10 = 4.4

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