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How can I calculate the following integral :

$$\int_{0}^{s}{\lambda \cdot |t|^{q-2}\cdot t\text{dt}}$$

$q \geq6$, $\lambda >0$ and $ s \in \mathbb{R}$.

It's hard for me to calcualte this integral because I don't know what I have to do with the module.

Thanks :)

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you dont have to do anything with the module since you are integrating from 0 to 8. Just open it positively. – mathemagician Jan 11 '13 at 8:22
@mathemagician why from $0$ to $8$. The heads of the integral are $0$ and $s \in \mathbb{R}$. – Iuli Jan 11 '13 at 8:23
$s \in \mathbb{R}$ or should $s$ be $\geqslant0$? Otherwise the integral doesn't quite make sense. If $s \geqslant0$, then you can remove then absolute value. – thang Jan 11 '13 at 8:26
then just break it into two cases. s can only be either $\geqslant 0$ or $< 0$, but not both. in each of these case, you can deal with the absolute value appropriately. – thang Jan 11 '13 at 8:30
Why don't you at least try to do it yourself? – Did Jan 11 '13 at 8:45
up vote 3 down vote accepted

It is best to discuss this problem by distinguishing two cases:

  • $s>0$: then $0<t<s$ so that $t$ is always positive. For $t>0$, we have $|t|=t$ and thus $$\int_{0}^{s}\lambda |t|^{q-2} t \,dt = \int_{0}^{s}\lambda t^{q-2} t \,dt = \lambda \int_{0}^{s} t^{q-1} \,dt = \frac{\lambda s^{q}}{q}.$$

  • $s<0$: then $s<t<0$ so that $t$ is always negative. For $t<0$, we have $|t|=-t$ and thus $$\int_{0}^{s}\lambda |t|^{q-2} t \,dt = \int_{0}^{s}\lambda (-t)^{q-2} t \,dt = \lambda \underbrace{(-1)^{q-2}}_{=(-1)^q}\int_{0}^{s} t^{q-1} \,dt = (-1)^q\frac{\lambda s^{q}}{q}.$$

You can write these two cases in a single formula $$\int_{0}^{s}\lambda |t|^{q-2} t \,dt = \frac{\lambda |s|^{q}}{q}$$

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