DefineG(x) =Z x20g(t)dt, x ∈ [0, 1].Prove that G is differentiable on [0, 1]. Compute G′(0) and G′(1)

To prove that G is differentiable on [0, 1], we need to show that the derivative of G exists for all x in [0, 1].

The derivative of G at a point x in [0, 1] is given by the limit as h approaches 0 of [G(x+h) - G(x)]/h.

G(x) is defined as the integral from 0 to x of 20g(t)dt. So, G(x+h) is the integral from 0 to x+h of 20g(t)dt.

The difference G(x+h) - G(x) is then the integral from x to x+h of 20g(t)dt.

As h approaches 0, this integral approaches 20g(x)*h.

So, the limit as h approaches 0 of [G(x+h) - G(x)]/h is 20g(x), which exists for all x in [0, 1] because g is a continuous function on [0, 1].

Therefore, G is differentiable on [0, 1], and its derivative G' is given by 20g(x).

To compute G'(0) and G'(1), we simply evaluate 20g(x) at x=0 and x=1.

So, G'(0) = 20g(0) and G'(1) = 20g(1).