Science:Math Exam Resources/Courses/MATH101 B/April 2025/Question 01 (b)/Solution 1

By examining the sign of $k(x)$ on the intervals $(0,3)$, $(3,5)$, and $(5,8)$, we have that $\underset{0}{\overset{3}{\int }}k(x)dx>0$, $\underset{3}{\overset{5}{\int }}k(x)dx<0$, and $\underset{5}{\overset{8}{\int }}k(x)dx<0$. Hence, $$\begin{array}{r}\underset{0}{\overset{3}{\int }}k(x)dx>\underset{0}{\overset{3}{\int }}k(x)dx+\underset{3}{\overset{5}{\int }}k(x)dx>\underset{0}{\overset{3}{\int }}k(x)dx+\underset{3}{\overset{5}{\int }}k(x)dx+\underset{5}{\overset{8}{\int }}k(x)dx,\end{array}$$ so $C<B<A$.