Sr Examen
Integral de t*exp(t)*exp(-p*t) dt
Solución
Ha introducido
[src]
oo / | | t -p*t | t*e *e dt | / 0
$$\int\limits_{0}^{\infty} t e^{t} e^{- p t}\, dt$$
Integral((t*exp(t))*exp((-p)*t), (t, 0, oo))
Respuesta (Indefinida)
[src]
// 2 \
|| t |
|| -- for p = 1|
|| 2 |
/ || | // t for -1 + p = 0\
| ||/ t*(1 - p) | || |
| t -p*t ||| e 2 | || t*(1 - p) |
| t*e *e dt = C - |<|------------ for 1 + p - 2*p != 0 | + t*|<-e |
| ||| 2 | ||------------ otherwise |
/ ||<1 + p - 2*p otherwise| || -1 + p |
||| | \\ /
||| -t |
||| ------ otherwise |
||\ -1 + p |
\\ /
$$\int t e^{t} e^{- p t}\, dt = C + t \left(\begin{cases} t & \text{for}\: p - 1 = 0 \\- \frac{e^{t \left(1 - p\right)}}{p - 1} & \text{otherwise} \end{cases}\right) - \begin{cases} \frac{t^{2}}{2} & \text{for}\: p = 1 \\\begin{cases} \frac{e^{t \left(1 - p\right)}}{p^{2} - 2 p + 1} & \text{for}\: p^{2} - 2 p + 1 \neq 0 \\- \frac{t}{p - 1} & \text{otherwise} \end{cases} & \text{otherwise} \end{cases}$$
Respuesta
[src]
/ 1 / pi 1 \ | -------- for And|-1 + |p|*cos(|arg(p)|) > 0, |arg(p)| < --, - != 1| | 2 \ 2 p / | (1 - p) | | oo < / | | | | t -p*t | | t*e *e dt otherwise | | |/ \0
$$\begin{cases} \frac{1}{\left(1 - p\right)^{2}} & \text{for}\: \cos{\left(\left|{\arg{\left(p \right)}}\right| \right)} \left|{p}\right| - 1 > 0 \wedge \left|{\arg{\left(p \right)}}\right| < \frac{\pi}{2} \wedge \frac{1}{p} \neq 1 \\\int\limits_{0}^{\infty} t e^{t} e^{- p t}\, dt & \text{otherwise} \end{cases}$$
=
=
/ 1 / pi 1 \ | -------- for And|-1 + |p|*cos(|arg(p)|) > 0, |arg(p)| < --, - != 1| | 2 \ 2 p / | (1 - p) | | oo < / | | | | t -p*t | | t*e *e dt otherwise | | |/ \0
$$\begin{cases} \frac{1}{\left(1 - p\right)^{2}} & \text{for}\: \cos{\left(\left|{\arg{\left(p \right)}}\right| \right)} \left|{p}\right| - 1 > 0 \wedge \left|{\arg{\left(p \right)}}\right| < \frac{\pi}{2} \wedge \frac{1}{p} \neq 1 \\\int\limits_{0}^{\infty} t e^{t} e^{- p t}\, dt & \text{otherwise} \end{cases}$$
Piecewise(((1 - p)^(-2), (Ne(1/p, 1))∧(Abs(arg(p)) < pi/2)∧(-1 + |p|*cos(Abs(arg(p))) > 0)), (Integral(t*exp(t)*exp(-p*t), (t, 0, oo)), True))
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.