Question

Evaluate ${\int }_0^2|x^2+2x-3|dx$

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (C)

4

$I={\int }_0^2|x^2+2x-3|dx={\int }_0^1|x^2+2x-3|dx+{\int }_1^2|x^2+2x-3|dx$
Therefore
$I=-{\int }_0^1(x^2+2x-3)dx+{\int }_1^2(x^2+2x-3)dx={-[\frac{x^3}{3}+x^2-3x]}_0^1+{[\frac{x^3}{3}+x^2-3x]}_1^2=4$