Evaluate `int_(0)^(2)7^xdx`

Answer

A, B

Explanation

`int_(0)^(x)[x]dx=int_(0)^(1)0dx+int_(1)^(2)1dx+int_(2)^(3)2dx+…………….`
`+int_([x]-1)^([x])([x]-1)dx+int_([x])^(x)[x]dx`
`=0+1+2+3+……….+([x]-1)+[x](x-[x])`
`=(([x]-1)[x])/2+[x]{x}`…………..1
Now `int_(0)^([x]) dx=[(x^(2))/2]_(0)^([x])=([x]^(2))/2`..............2
Comparing 1 and 2 we get
`([x]-1)[x])/2+[x]{x}=([x]^(2))/2`
`:.[x]^(2)-[x]+2[x]{x}=[x]^(2)`
`=[x]=0` or `{x}=1//2`