NIMCET 2014 Question Paper with solutions

We have
l = $\underset{1}{\overset{9}{\int }}$ [x] dx
= $\underset{1}{\overset{2}{\int }}$ 1 dx + $\underset{2}{\overset{3}{\int }}$ 2 dx + $\underset{3}{\overset{4}{\int }}$ 3 dx + $\underset{4}{\overset{3}{\int }}$ 4 dx + $\underset{5}{\overset{6}{\int }}$ 5 dx + $\underset{6}{\overset{7}{\int }}$ 6 dx + $\underset{7}{\overset{8}{\int }}$ 7 dx + $\underset{8}{\overset{9}{\int }}$ 8 dx
= (2-1) +2(3-2)+ 3(4-3)+ 4(5-4)+ 5(6-5) + 6(7—6) + 7(8-7) + 8(9-8)
= 1 + 2 + ... + 8
= $\frac{8\times 9}{2}$ = 36