Question

${\int }_1^4|x^2-5x+6|dx$

Answer 1

${\int }_1^4|x^2-5x+6|dx$

$=\frac{(x^2-5x+6)}{\left|\right(x^2-5x+6\left)\right|}\int (x^2-5x+6)dx$

$=\frac{(x^2-5x+6)}{\left|\right(x^2-5x+6\left)\right|}(\frac{x^3}{3}-5\frac{x^2}{2}+6x)$

$\int |x^2-5x+6|dx=\frac{(x-3)(x-2)}{\left|\right(x-3\left)\left|\right|\right(x-2\left)\right|}(\frac{x^3}{3}-5\frac{x^2}{2}+6x)+C$

${\int }_1^4|x^2-5x+6|dx=\frac{(x-3)(x-2)}{\left|\right(x-3\left)\left|\right|\right(x-2\left)\right|}(\frac{x^3}{3}-5\frac{x^2}{2}+6x)+C$

$=\frac{(4-3)(4-2)}{\left|\right(4-3\left)\left|\right|\right(4-2\left)\right|}(\frac{4^3}{3}-5\frac{4^2}{2}+6\times 4)-\frac{(1-3)(1-2)}{\left|\right(1-3\left)\left|\right|\right(1-2\left)\right|}(\frac{1^3}{3}-5\frac{1^2}{2}+6\times 1)$

$=(\frac{4^3}{3}-5\frac{4^2}{2}+6\times 4)-(\frac{1^3}{3}-5\frac{1^2}{2}+6\times 1)$

$=\frac{126-225+108}{6}$

$=\frac{3}{2}$

${\int }_1^4|x^2-5x+6|dx=\frac{3}{2}$