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\documentclass{article} \usepackage{amsmath} \begin{document} (****) (a) Find the indefinite integrals below, noting that the expression $e^{-2x} (A\cos(mx) + B\sin(mx))$ might be relevant: \begin{align} \int e^{-2x} \sin(mx) dx &= \\ \int e^{-2x} \cos(mx) dx &= \\ \end{align} (b) A random variable $Y$ has an exponential distribution, with the probability density shown here: $$p(y) = \begin{cases} 2e^{-2y}, & \text{for } y \ge 0, \\ 0, & \text{for } y < 0. \end{cases}$$ Find the expected value of the random variable $X = 6 \cos(6Y) + 6 \sin(6Y)$. $E[X] = E[6 \cos(6Y) + 6 \sin(6Y)] = $ Hint: For any given random variable $Y$ and function $h$, $E[h(Y)] = \int_{-\infty}^{\infty} h(y)p(y) dy$. \end{document}

          \documentclass{article}
\usepackage{amsmath}
\begin{document}
(****) (a) Find the indefinite integrals below, noting that the expression $e^{-2x} (A\cos(mx) + B\sin(mx))$ might be relevant:
\begin{align*}
\int e^{-2x} \sin(mx) dx &= \\
\int e^{-2x} \cos(mx) dx &= \\
\end{align*}
(b) A random variable $Y$ has an exponential distribution, with the probability density shown here:
$$p(y) = \begin{cases} 2e^{-2y}, & \text{for } y \ge 0, \\ 0, & \text{for } y < 0. \end{cases}$$
Find the expected value of the random variable $X = 6 \cos(6Y) + 6 \sin(6Y)$.
$E[X] = E[6 \cos(6Y) + 6 \sin(6Y)] = $
Hint: For any given random variable $Y$ and function $h$, $E[h(Y)] = \int_{-\infty}^{\infty} h(y)p(y) dy$.
\end{document}

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(****) (a) Find the indefinite integrals below, noting that the expression e^-2x (Acos(mx) + Bsin(mx)) might be relevant:

∫ e^-2xsin(mx) dx =
∫ e^-2xcos(mx) dx =

(b) A random variable Y has an exponential distribution, with the probability density shown here:

p(y) = 2e^-2y, for y ≥ 0,
0, for y < 0.

Find the expected value of the random variable X = 6 cos(6Y) + 6 sin(6Y).
E[X] = E[6 cos(6Y) + 6 sin(6Y)] =
Hint: For any given random variable Y and function h, E[h(Y)] = ∫-∞^∞ h(y)p(y) dy.

Added by Jennifer O.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

06/15/2023

Step 1

Step 1: Find the indefinite integral of e^(-2x)sin(mx)dx: Integrate by parts: Let u = sin(mx) and dv = e^(-2x)dx Then du = mcos(mx)dx and v = -1/2e^(-2x) Using the formula for integration by parts: ∫ e^(-2x)sin(mx)dx = -1/2e^(-2x)sin(mx) - ∫ Show more…

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(******darrx) (a) Find the indefinite integrals below, noting that the expression e^(-2x)(Acos(mx)+Bsin(mx)) might be relevant: Answer: ∫ e^(-2x)sin(mx)dx= Answer: ∫ e^(-2x)cos(mx)dx= (b) A random variable Y has an exponential distribution, with the probability density shown here: p(y)={(2e^(-2y), for y>=0,),(0, for y<0.):} Find the expected value of the random variable X=6cos(6Y)+6sin(6Y). Answer: E[X]=E[6cos(6Y)+6sin(6Y)]= Hint: For any given random variable Y and function h, E[h(Y)]=∫ h(y)p(y)dy. (**) (a) Find the indefinite integrals below, noting that the expression e^(-2) (Acos(m) + Bsin(m)) might be relevant: Answer: -2x sin(mx)dx= Answer: -2xcos(mx)dx= (b) A random variable Y has an exponential distribution, with the probability density shown here: (2e^(-2y) for y>=0, p(y) = 0 for y < 0. Find the expected value of the random variable X=6cos(6Y)+6sin(6Y). Answer: E[X]=E[6cos(6Y)+6sin(6Y)]= Hint: For any given random variable Y and function h, E[h(Y)]=∫ h(y)p(y)dy.

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