Question

Calculate ${\int }_{-2}^{3}f\left(x\right)dx$ where
$f\left(x\right)=\{\begin{array}{cc}6& ifx>1\\ 3x^2& ifx\le 1\end{array}$

• A. 19
• B. 21
• C. 17
• D. 15

Answer 1

The correct option is B 21


Here, we can see that the integrand f(x) in not continuous in the interval (-2, 3) as it has a discontinuity at x = 1. So we can’t just integrate f(x) and put limits.We’ll have to break this integral into integrals which have limits such that integrands is continuous in those limits.
${\int }_{-2}^{3}f\left(x\right)dx={\int }_{-2}^{1}3x^{2}dx+{\int }_1^{3}6dx$
Now we can integrate these integrands and put limits.
${\int }_{-2}^{1}3x^{2}dx+{\int }_1^{3}6dx=\left(x^3\right){|}_{-2}^1+\left(6x\right){|}_1^3=1-(-8)+18-6$
= 21