Question

Suppose \(|f(x)|\leq M\) for all \(x\in[a,b]\). For \(x\in[a,b]\) set \(F(x)=\int_a^x f(t)dt\). Show that F(x) is continuous on [a,b]
Please, help

Answer 1

@GIL.ojei

Answer 2

for a function to be continues, what must he posses?

Answer 3

\(lim_{x\rightarrow x_0} f(x) = f(x_0)\)

Answer 4

very good. it must posses a limit. or its limit must exist

Answer 5

for all x_0 on [a,b]

Answer 6

yap

Answer 7

then?

Answer 8

\[F(x)=\int\limits_a^x f(t)dt \] that is the function. integrate it with respect to t and subtitute a and x and conclude

Answer 9

using \[ |f(x)|≤M for all x∈[a,b] \] to solve

Answer 10

Got you

Answer 11

I think of Riemann integrable of F(x) to show that F(x) integrable, hence it is continuous. But it seems I complicate the problem

Answer 12

hmmm

Answer 13

ok, let finish it. :) to make sure I don't overestimate my ability. hehehe

Answer 14

can you intigrate\[F(x)=\int\limits_a^x f(t)dt \]

Answer 15

why not? it just f(x) - f(a) right?

Answer 16

and f(x) - f(a) < f(x)
|f(x) -f(a) |<|f(x) < M

Answer 17

I will show my stuff here. If you can, check when you are back, please. Thanks for the help though

Answer 18

u have to use the definition of continuity:
\[\large (\forall\varepsilon>0)(\exists\delta>0)(\forall x)(|x-x_0|<\delta\Rightarrow
|F(x)-F(x_0)|<\varepsilon ) \]

Answer 19

Let \(c\in[a.b]\). If \(h>0\) then
\[\large F(c+h)-F(c)=\int_a^{c+h}f-\int_a^cf=\int_c^{c+h}f. \]

Answer 20

Since \(-M\leq f(x)\leq M\), then
\[ -M\cdot h\leq\int_c^{c+h}f\leq M\cdot h \]
in other words
\[ -M\cdot h\leq F(c+h)-F(c)\leq M\cdot h.\tag{1} \]
If \(h<0\) u can get a similar inequality.

Answer 21

Notice that
\[ F(c+h)-F(c)=\int_c^{c+h}f=-\int_{c+h}^cf. \]
Then, over the interval \([c+h,c]\) of length \(-h\) we have
\[ M\cdot h\leq\int_{c+h}^cf\leq-M\cdot h \]
which when multiplied by \(-1\) becomes
\[ M\cdot h\leq F(c+h)-F(c)\leq-M\cdot h\tag{2} \]

Answer 22

(1) and (2) can be combined into
\[ |F(c+h)-F(c)|\leq M\cdot|h|. \]
Therefore, if \(\varepsilon>0\), we have that
\[ |F(c+h)-F(c)|<\varepsilon, \]
whenever \(|h|<\varepsilon/M\). This proves that
\[ \lim_{h\to0}F(c+h)=F(c) \]
which means that \(F\) is continous at \(c\).

Answer 23

now this is very corrct. thanks @helder_edwin

Answer 24

Thanks for the help. Much appreciate. @GIL.ojei @helder-edwin