DefineG(x) =Z x20g(t)dt, x ∈ [0, 1].Prove that G is differentiable on [0, 1]. Compute G′(0) and G′(1)
Question
DefineG(x) =Z x20g(t)dt, x ∈ [0, 1].Prove that G is differentiable on [0, 1]. Compute G′(0) and G′(1)
Solution
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.
Therefore, [G(x+h) - G(x)]/h = [∫_0^(x+h) 20g(t)dt - ∫_0^x 20g(t)dt] / h = ∫_x^(x+h) 20g(t)dt / h.
As h approaches 0, the integral ∫_x^(x+h) 20g(t)dt approaches 20g(x)h, so the limit 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 G'(x) = 20g(x) for all x in [0, 1].
To compute G'(0) and G'(1), we simply substitute x = 0 and x = 1 into the formula for G':
G'(0) = 20g(0) G'(1) = 20g(1)
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