Answered: actuaZ D X=Ytmacking weigh…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Attached is a screen shot from my professor's working a textbook problem on the topic: Cumulative Distribution Functionsand expected Values

Can you explain why he suddenly changes the upper and lower bounds of the integral from 2-4 to -1 to 1? I'm rusty on my integral calculus. Thank you.

actuaZ
D X=Ytmacking weigh
Şk.(i-cx-3)³), ifɔ
147
f(x)=
0, otherwise
(4) k=? SFxsdx=1)
→1= (k(1-(x-3)²)dx =k ((1-t³)dt-
t=x-3
dt =(x-3)'dx=dx
3
|t=1
3 14=-1
1=k. (를+를) 1=k.=k-류
4
(c) P(X>3)= Sfcx)dx= 3(1-(x-3)° d«
t=x-3

Expert Solution

Step 1

Given, the PMF, $f (x) = \{\begin{cases} k (1 - {(x - 3)}^{2}), 2 \leq x \leq 4 \\ 0, o t h e r w i s e \end{cases}$

Step 2

(b) To find k:

From the property of PMF $\int_{- \infty}^{\infty} f (x) d x = 1$ .

Consider,

$\int_{- \infty}^{\infty} f (x) d x = 1 \Rightarrow \int_{- \infty}^{2} 0 d x + \int_{2}^{4} k (1 - {(x - 3)}^{2}) d x + \int_{4}^{\infty} 0 d x = 1 \Rightarrow \int_{2}^{4} k (1 - {(x - 3)}^{2}) d x = 1 S u b s t i t u t e, x - 3 = t \Rightarrow d x = d t & x v a r i e s f r o m 2 t o 4 \Rightarrow t = x - 3 v a r i e s f r o m t = 2 - 3 = - 1 t o t = 4 - 3 = 1 . T h u s t h e i n t g r a l b e c o m e s, \int_{- 1}^{1} k (1 - t^{2}) d t = 1 \Rightarrow k \int_{- 1}^{1} (1 - t^{2}) d t = 1 \Rightarrow k {[t - \frac{t^{3}}{3}]}_{- 1}^{1} = 1 \Rightarrow k [(1 - \frac{1}{3}) - ((- 1) - (\frac{{(- 1)}^{3}}{3}))] = 1 \Rightarrow k [\frac{2}{3} + \frac{2}{3}] = 1 \Rightarrow k \times \frac{4}{3} = 1 \Rightarrow k = \frac{3}{4}$

Thus, $k = \frac{3}{4}$ .

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