Question

Evaluate:

${\int }_0^1(x^3-{4x}^2+3x)dx-{\int }_2^3(x^3-4x^2+3x)dx$

Answer 1

Given,

${\int }_0^1{x}^3-4x^2+3x-{\int }_2^3{x}^3-4x^2+3xdx$

$={\int }_0^1{x}^{3}dx-{\int }_0^{1}4x^{2}dx+{\int }_0^{1}3xdx-[{\int }_2^3{x}^{3}dx-{\int }_2^{3}4x^{2}dx+{\int }_2^{3}3xdx]$

$=\frac{1}{4}-\frac{4}{3}+\frac{3}{2}-[\frac{65}{4}-\frac{76}{3}+\frac{15}{2}]$

$=\frac{5}{12}-[-\frac{19}{12}]$

$=\frac{5}{12}+\frac{19}{12}$

$=\frac{24}{12}$

$=2$