Let f : [a, b] → [1, \infty) be a continuous function and let g : R → R be defined as g(x) = \left{\right. 0 , if \textrm{ } x < a \\ \int_{a}^{x} f \left(\right. t \left.\right) \textrm{ } dt , if \textrm{ } a \leq x \leq b \\ \int_{a}^{b} f \left(\right. t \left.\right) \textrm{ } dt , if \textrm{ } x

Engineering

Mathematics

Properties of Definite Integral

Question

Let f : [a, b] → [1, $\infty$ ) be a continuous function and let g : R → R be defined as

g(x) = ${\begin{cases} 0, if x < a \\ \int_{a}^{x} f (t) dt, if a \leq x \leq b \\ \int_{a}^{b} f (t) dt, if x > b \end{cases}$

Then

g(x) is continuous but not differentiable at a.

g(x) is continuous and not differentiable at b.

g(x) is continuous and differentiable at either a or b but not both.

g(x) is differentiable on R.

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Checking differentiability at x = a and x = b

g'(a⁺) = $\underset{h \to 0}{Lim} (\frac{g (a + h) - g (a)}{h}) = \underset{h \to 0}{Lim} (\frac{\int_{a}^{a + h} f (t) dt - 0}{h}) = \underset{h \to 0}{Lim} \frac{f (a + h)}{1} \Rightarrow$ g'(a⁺) = f(a)

g'(a^–) = $\underset{h \to 0}{Lim} \frac{g (a - h) - g (a)}{- h} = \underset{h \to 0}{Lim} \frac{0 - 0}{- h} = 0$

|||l^y g'(b⁺) = 0 and g'(b^–) = f(b)

But f(a) or f(b) $\neq$ 0 as co‑domain of f(x) $\in$ [1, $\infty$ )

$∴$ non‑derivable at x = a and b.