Question

Calcule o valor das integrais indefinidas:
b) raiz de (3 - 2x)² ds

Answer 1

$\int\ { \sqrt{(3-2x)^2} } \, ds = \int\ {3-2x } \, ds=s(3-2x)=3s-2xs$

No caso de s no lugar de x:

$\int\ { \sqrt{(3-2s)^2} } \, ds = \int\ {3-2s } \, ds=3s-\frac{2s^2}{2}+C=3s-s^2+C$

Answer 2

$\displaystyle\int{\sqrt{(3-2s)^{2}}\,ds}\\ \\ \\ =\int{|3-2s|\,ds}\\ \\ \\ =\int{|2s-3|\,ds}~~~~~~\mathbf{(i)}$


Mudança de variável:

$2s-3=u~\Rightarrow~2\,ds=du~\Rightarrow~ds=\dfrac{1}{2}\,du$


Substituindo, a integral $\mathbf{(i)}$ fica

$\displaystyle\int{|u|\cdot \dfrac{1}{2}\,du}\\ \\ \\ =\dfrac{1}{2}\int{|u|\,du}=\left\{ \begin{array}{ll} \dfrac{1}{2}\displaystyle\int{u\,du}\,,&\text{se }u>0\\ \\ \dfrac{1}{2}\displaystyle\int{(-u)\,du}\,,&\text{se }u<0 \end{array} \right.\\ \\ \\ \\ =\left\{ \begin{array}{ll} \dfrac{u^{2}}{4}\,,&\text{se }u>0\\ \\ -\dfrac{u^{2}}{4}\,,&\text{se }u<0 \end{array}\right.\\ \\ \\ =\dfrac{u}{|u|}\cdot \dfrac{u^{2}}{4}+C\\ \\ \\ =\dfrac{u^{3}}{4|u|}+C$

$=\dfrac{(2s-3)^{3}}{4\cdot |2s-3|}+C\\ \\ \\ =-\dfrac{(3-2s)^{3}}{4\cdot |3-2s|}+C\\ \\ \\ =-\dfrac{(3-2s)^{3}}{4\sqrt{(3-2s)^{2}}}+C$

$\Rightarrow~\boxed{\begin{array}{c} \displaystyle\int{\sqrt{(3-2s)^{2}}\,ds}=-\dfrac{(3-2s)^{3}}{4\sqrt{(3-2s)^{2}}}+C \end{array}}$