Answer to Question #91395 in Calculus for Sajid

Question

Q. choose the correct option.
Q. Let H(x)=∫_a^x▒f(t)dt, for a≤x≤b,then which of the following is true?
(i) H may not be continuous on [a,b]
(ii) f is continuous at c∈[a,b],then H is differentiable at c.
(iii) H is continuous on [a,b].
a.(i) only
b.(i) and (ii) only
c.(ii) and (iii) only.
d.(iii) only

Answer 1

The correct answer is b) (i) and (ii) only.

I) If f is not integrable on [a, b] (for example, if f is a Dirichlet function), then H will not be continuous on [a, b]. A Dirichlet function is not Riemann integrable (the upper limit sum=1 and the lower limit sum=0), hence in this case $\int_{a}^{x}f(t)dt$ doesn't exist.

II) If f is continuous at c∈[a, b], then

$$\frac{H(c+h)-H(c)}{h}=\frac{1}{h}\left(\int\limits_a^{c+h}f(t)dt-\int\limits_a^{c}f(t)dt\right)=\frac{1}{h}\int\limits_{c}^{c+h}f(t)dt$$

Assuming that

$$\int\limits_{c}^{c+h}f(c)dt=hf(c)$$

We will have

$$\left|\frac{H(c+h)-H(c)}{h}-f(c)\right|=\left|\frac{1}{h}\int\limits_{c}^{c+h}f(t)dt-f(c)\right|=\left|\frac{1}{h}\int\limits_{c}^{c+h}(f(t)-f(c))dt\right|$$

Since f is continuous at c

$$\lim_{t\to c}\left(f(t)-f(c)\right)=0$$

Therefore

$$\forall\varepsilon>0 \exist \delta>0, \forall t: |t-c|\le\delta \implies |f(t)-f(c)|\le \varepsilon$$

Hence if

$$|h|<\delta$$

$$\left|\frac{1}{h}\int\limits_{c}^{c+h}(f(t)-f(c))dt\right|\le\left|\frac{1}{h}\right|\left|\int\limits_{c}^{c+h}\varepsilon dt\right|=\left|\frac{1}{h}\right|\varepsilon\left|h\right|=\varepsilon$$

which means that if $h \to 0$

$$\left|\frac{1}{h}\int\limits_{c}^{c+h}(f(t)-f(c))dt\right|\to 0$$

Thus, H is differentiable at c