Sr Examen
Integral de k^2*t^2*exp(t*(-k))/3 dt
Solución
Ha introducido
[src]
oo / | | 2 2 t*(-k) | k *t *e | ------------- dt | 3 | / 0
$$\int\limits_{0}^{\infty} \frac{k^{2} t^{2} e^{- k t}}{3}\, dt$$
Integral(((k^2*t^2)*exp(t*(-k)))/3, (t, 0, oo))
Respuesta (Indefinida)
[src]
/ // 3 \ \
| || t | |
| || -- for k = 0| |
| || 3 | |
| || | |
| ||/ -k*t | // t for k = 0\|
| |||(1 + k*t)*e 3 | || ||
2 | |||--------------- for k != 0 | 2 || -k*t ||
k *|- 2*|<| 3 | + t *|<-e ||
| ||| k | ||------- otherwise||
| ||< otherwise| || k ||
| ||| 2 | \\ /|
| ||| -t | |
/ | ||| ---- otherwise | |
| | ||| 2*k | |
| 2 2 t*(-k) | ||\ | |
| k *t *e \ \\ / /
| ------------- dt = C + -------------------------------------------------------------------------------
| 3 3
|
/
$$\int \frac{k^{2} t^{2} e^{- k t}}{3}\, dt = C + \frac{k^{2} \left(t^{2} \left(\begin{cases} t & \text{for}\: k = 0 \\- \frac{e^{- k t}}{k} & \text{otherwise} \end{cases}\right) - 2 \left(\begin{cases} \frac{t^{3}}{3} & \text{for}\: k = 0 \\\begin{cases} \frac{\left(k t + 1\right) e^{- k t}}{k^{3}} & \text{for}\: k^{3} \neq 0 \\- \frac{t^{2}}{2 k} & \text{otherwise} \end{cases} & \text{otherwise} \end{cases}\right)\right)}{3}$$
Respuesta
[src]
/ 2 pi | --- for |arg(k)| < -- | 3*k 2 | | oo | / | | < | 2 2 -k*t | | k *t *e | | ----------- dt otherwise | | 3 | | |/ |0 \
$$\begin{cases} \frac{2}{3 k} & \text{for}\: \left|{\arg{\left(k \right)}}\right| < \frac{\pi}{2} \\\int\limits_{0}^{\infty} \frac{k^{2} t^{2} e^{- k t}}{3}\, dt & \text{otherwise} \end{cases}$$
=
=
/ 2 pi | --- for |arg(k)| < -- | 3*k 2 | | oo | / | | < | 2 2 -k*t | | k *t *e | | ----------- dt otherwise | | 3 | | |/ |0 \
$$\begin{cases} \frac{2}{3 k} & \text{for}\: \left|{\arg{\left(k \right)}}\right| < \frac{\pi}{2} \\\int\limits_{0}^{\infty} \frac{k^{2} t^{2} e^{- k t}}{3}\, dt & \text{otherwise} \end{cases}$$
Piecewise((2/(3*k), Abs(arg(k)) < pi/2), (Integral(k^2*t^2*exp(-k*t)/3, (t, 0, oo)), True))
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.