Evaluate `int_(0)^(1) 6^x dx`

Answer

A, C, D

Explanation

(1) The expression `f(x)f(c)AAx epsilon(c-h,c+h)` where `hto0^(+)` is equivalent to `lim_(xtoc)f(x)f(c)`which is equal to `f(c))^(2)` because `f(x)` is continuous.
There `f(x)f(c)gt0AA(c-h,c+h)` where `hto0^(+)`.
(2) We have `I=lim_(nto oo) 1n In[(1+1/n)(1+2/n)........(1+n/n)]`
`=lim_(nto oo) 1/n In prod_(k=1)^(n)(1+k/n)`
`=lim_(nto oo) 1/n sum_(k=1)^(n)In(1+k/n)`
`=int_(0)^(1)log_(e)(1+x)dx`
`=int_(1)^(2)Inx dx=[(Inx-1)]_(1)^(2)=-1+2In2` ltbgt (3) Given `f(x)gt0` or `int_(a)^(b)f(x)dxge0`.
But given `int_(a)^(b)f(x)dx=0`. So, this can be true only when `f(x)=0`
(4) `int_(a)^(b)f(x)dx+0` i.e. `y=f(x)` cuts x-axis at least once.