Question

integrate 1 to 3 abs(x-1) + abs(x-2)​

Answer 1

$\displaystyle\sf \int_{0}^{3} |x-1|+|x-2| dx$

rewrite it as ↓

$\displaystyle\sf \int_{0}^{3} |x-2|+|x-1| dx$

integrate the sum term by term:

$\displaystyle\sf \int_{0}^{3} |x-1| dx + \displaystyle\sf \int_{0}^{3} |x-2| dx$

for the integrand |x-1|, substitute$\sf u = x -1$ and du = dx.

this gives a new lower limit u = 0-1 = -1 and upper limit u = 3-1 = 2

$\displaystyle\sf \int_{-1}^{2}|u| dx + \displaystyle\sf \int_{0}^{3} |x-2| dx$

the only zero of u in the integration domain of u = 0. factor out the absolute value.

$\displaystyle\sf \left| \dfrac{u^2}{2} \right|_{-1}^{0} \Bigg | + \left| \int_{0}^{2} u \ du \right| + \int_0^3 |x-2| dx$

$\displaystyle\sf = \dfrac{5}{2} + \int_0^3 |x-2| dx$

substituting |x-2| as s, dx = ds

$\displaystyle\sf \dfrac{9}{2} + \left| \int_0^3 s \ ds \right|$

$\sf = \dfrac{9}{2} + \dfrac{1}{2}$

$\sf = 5$

know more!

$\boxed{\boxed{\begin{minipage}{4cm}\displaystyle\circ\sf\:\int{1\:dx}=x+c\\\\\circ\sf\:\int{a\:dx}=ax+c\\\\\circ\sf\:\int{x^n\:dx}=\dfrac{x^{n+1}}{n+1}+c\\\\\circ\sf\:\int{sin\:x\:dx}=-cos\:x+c\\\\\circ\sf\:\int{cos\:x\:dx}=sin\:x+c\\\\\circ\sf\:\int{sec^2x\:dx}=tan\:x+c\\\\\circ\sf\:\int{e^x\:dx}=e^x+c\end{minipage}}}$

Answer 2

Answer:

Step-by-step explanation:

substituting |x-2| as s, dx = ds