Question

lf ${\int }_1^4|x-3|dx=2A+B$ then $A$ and $B$ can be

• A. $A=\frac{3}{2},B=4$
• B. $A=\frac{1}{2},B=1$
• C. $A=2,B=-\frac{3}{2}$
• D. $A=\frac{1}{2},B=\frac{3}{2}$

Answer 1

Answer: (C), (D)

The correct options are
C $A=2,B=-\frac{3}{2}$
D $A=\frac{1}{2},B=\frac{3}{2}$

${\int }_1^4|x-3|dx=2A+B$
$-{\int }_{+1}^3(x-3)dx+{\int }_3^4(x-3)dx$
$-\left(\frac{(x-3{)}^2}{2}\right){|}_{+1}^3+\frac{(x-3{)}^2}{2}{|}_3^4$
$-[0-2]+\left[1/2\right]$
$5/2$
from the options verify such that
$2A+B=5/2$
So, $A=2;B=-3/2$ $A=1/2;B=3/2$