Question

Evaluate: $I={\int }_1^2(2t-4)dt$

• A. $-1$
• B. $1$
• C. $2$
• D. $-2$

Answer 1

The correct option is A $-1$

$I={\int }_1^2(2t-4)dt=2{\int }_1^{2}tdt-4{\int }_1^{2}dt$

$⟹I=2{\left[\frac{t^2}{2}\right]}_1^2-4[t{]}_1^2$

$⟹I=(4-1)-4(2-1)$

$⟹I=-1$

Hence, answer is option-(A).