Question

G(x)=∫ tan(t)dt How to find G(1)= and How to find G′(π/6)= ? (a=1 & b=x)

Answer 1

`G(1) =0` and `G'(pi/6) = sqrt3/3`

Explanation:

`G(x) = int_1^x tant dt`

`G(1) = int_1^1 tant dt = 0`

Because, for every `f`, and every `a`, we have `int_a^a f(t)dt = 0`

`G'(x) = tanx`, so `G'(pi/6) = tan (pi/6) = 1/sqrt3 = sqrt3/3`