Suppose $f:Bbb{R}rightarrowBbb{R}$ is continuous. Define $G(x)=int_0^x(x-t)f(t)dt$ for all $xinBbb{R}$. Prove...
$begingroup$
This is a tricky example of the Fundamental Theorem of Calculus 1. I am looking for confirmation of my proof or answer to see if I am doing it correctly because this is different than the basic applications.
$underline{Proof}$
$G(x)=int_0^x(x-t)f(t)dtrightarrow G(x)=int_0^xxf(t)dt-int_0^xtf(t)dt$.
So, let $Q(x)=int_0^xf(t)dt$ and let $P(x)=int_0^xtf(t)dt$ then we have,
$G(x)=xQ(x)-P(x)$
$G'(x)=xQ'(x)+Q(x)-P'(x)=xf(x)+Q(x)-xf(x)=Q(x)$
hence,
$G''(x)=Q'(x)=f(x)$
∎
real-analysis calculus
asked Dec 6 '18 at 6:12
Albert DiazAlbert Diaz
925
$endgroup$
add a comment |
$begingroup$
This is a tricky example of the Fundamental Theorem of Calculus 1. I am looking for confirmation of my proof or answer to see if I am doing it correctly because this is different than the basic applications.
$underline{Proof}$
$G(x)=int_0^x(x-t)f(t)dtrightarrow G(x)=int_0^xxf(t)dt-int_0^xtf(t)dt$.
So, let $Q(x)=int_0^xf(t)dt$ and let $P(x)=int_0^xtf(t)dt$ then we have,
$G(x)=xQ(x)-P(x)$
$G'(x)=xQ'(x)+Q(x)-P'(x)=xf(x)+Q(x)-xf(x)=Q(x)$
hence,
$G''(x)=Q'(x)=f(x)$
∎
real-analysis calculus
asked Dec 6 '18 at 6:12
Albert DiazAlbert Diaz
925
$endgroup$
$begingroup$
Looks fine. You can also use the Leibniz integral rule.
$endgroup$
– Mattos
Dec 6 '18 at 6:15
$begingroup$
@Mattos, thank you I will also try this problem with Leibniz for practise
$endgroup$
– Albert Diaz
Dec 6 '18 at 6:22
add a comment |
$begingroup$
This is a tricky example of the Fundamental Theorem of Calculus 1. I am looking for confirmation of my proof or answer to see if I am doing it correctly because this is different than the basic applications.
$underline{Proof}$
$G(x)=int_0^x(x-t)f(t)dtrightarrow G(x)=int_0^xxf(t)dt-int_0^xtf(t)dt$.
So, let $Q(x)=int_0^xf(t)dt$ and let $P(x)=int_0^xtf(t)dt$ then we have,
$G(x)=xQ(x)-P(x)$
$G'(x)=xQ'(x)+Q(x)-P'(x)=xf(x)+Q(x)-xf(x)=Q(x)$
hence,
$G''(x)=Q'(x)=f(x)$
∎
real-analysis calculus
asked Dec 6 '18 at 6:12
Albert DiazAlbert Diaz
925
$endgroup$
This is a tricky example of the Fundamental Theorem of Calculus 1. I am looking for confirmation of my proof or answer to see if I am doing it correctly because this is different than the basic applications.
$underline{Proof}$
$G(x)=int_0^x(x-t)f(t)dtrightarrow G(x)=int_0^xxf(t)dt-int_0^xtf(t)dt$.
So, let $Q(x)=int_0^xf(t)dt$ and let $P(x)=int_0^xtf(t)dt$ then we have,
$G(x)=xQ(x)-P(x)$
$G'(x)=xQ'(x)+Q(x)-P'(x)=xf(x)+Q(x)-xf(x)=Q(x)$
hence,
$G''(x)=Q'(x)=f(x)$
∎
real-analysis calculus
real-analysis calculus
asked Dec 6 '18 at 6:12
Albert DiazAlbert Diaz
925
asked Dec 6 '18 at 6:12
Albert DiazAlbert Diaz
925
asked Dec 6 '18 at 6:12
Albert DiazAlbert Diaz
925
asked Dec 6 '18 at 6:12
Albert DiazAlbert Diaz
925
asked Dec 6 '18 at 6:12
Albert DiazAlbert Diaz
925
925
$begingroup$
Looks fine. You can also use the Leibniz integral rule.
$endgroup$
– Mattos
Dec 6 '18 at 6:15
$begingroup$
@Mattos, thank you I will also try this problem with Leibniz for practise
$endgroup$
– Albert Diaz
Dec 6 '18 at 6:22
add a comment |
$begingroup$
Looks fine. You can also use the Leibniz integral rule.
$endgroup$
– Mattos
Dec 6 '18 at 6:15
$begingroup$
@Mattos, thank you I will also try this problem with Leibniz for practise
$endgroup$
– Albert Diaz
Dec 6 '18 at 6:22
$begingroup$
Looks fine. You can also use the Leibniz integral rule.
$endgroup$
– Mattos
Dec 6 '18 at 6:15
$begingroup$
Looks fine. You can also use the Leibniz integral rule.
$endgroup$
– Mattos
Dec 6 '18 at 6:15
$begingroup$
@Mattos, thank you I will also try this problem with Leibniz for practise
$endgroup$
– Albert Diaz
Dec 6 '18 at 6:22
$begingroup$
@Mattos, thank you I will also try this problem with Leibniz for practise
$endgroup$
– Albert Diaz
Dec 6 '18 at 6:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your proof is correct. An alternative is to use integration by parts. With
$$
f_1(x) = int_0^x f(t) , dt , , quad f_2(x) = int_0^x f_1(t) , dt
$$
we have
$$
G(x) = bigllbrack (x-t)f_1(t) bigrrbrack_{t=0}^{t=x} +
int_0^x f_1(t) , dt = f_2(x)
$$
and therefore $G'' = f_2'' = f_1' = f$.
This generalizes to higher anti-derivatives:
$$
int_0^x frac{(x-t)^{n-1}}{(n-1)!} f(t) , dt
= int_0^x int_0^{t_1} cdots int_0^{t_{n-1}} f(t_n) , dt_n cdots dt_1
$$
for $n ge 1$.
answered Dec 6 '18 at 6:53
Martin RMartin R
27.9k33255
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
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$begingroup$
Your proof is correct. An alternative is to use integration by parts. With
$$
f_1(x) = int_0^x f(t) , dt , , quad f_2(x) = int_0^x f_1(t) , dt
$$
we have
$$
G(x) = bigllbrack (x-t)f_1(t) bigrrbrack_{t=0}^{t=x} +
int_0^x f_1(t) , dt = f_2(x)
$$
and therefore $G'' = f_2'' = f_1' = f$.
This generalizes to higher anti-derivatives:
$$
int_0^x frac{(x-t)^{n-1}}{(n-1)!} f(t) , dt
= int_0^x int_0^{t_1} cdots int_0^{t_{n-1}} f(t_n) , dt_n cdots dt_1
$$
for $n ge 1$.
answered Dec 6 '18 at 6:53
Martin RMartin R
27.9k33255
$endgroup$
add a comment |
$begingroup$
Your proof is correct. An alternative is to use integration by parts. With
$$
f_1(x) = int_0^x f(t) , dt , , quad f_2(x) = int_0^x f_1(t) , dt
$$
we have
$$
G(x) = bigllbrack (x-t)f_1(t) bigrrbrack_{t=0}^{t=x} +
int_0^x f_1(t) , dt = f_2(x)
$$
and therefore $G'' = f_2'' = f_1' = f$.
This generalizes to higher anti-derivatives:
$$
int_0^x frac{(x-t)^{n-1}}{(n-1)!} f(t) , dt
= int_0^x int_0^{t_1} cdots int_0^{t_{n-1}} f(t_n) , dt_n cdots dt_1
$$
for $n ge 1$.
answered Dec 6 '18 at 6:53
Martin RMartin R
27.9k33255
$endgroup$
add a comment |
$begingroup$
Your proof is correct. An alternative is to use integration by parts. With
$$
f_1(x) = int_0^x f(t) , dt , , quad f_2(x) = int_0^x f_1(t) , dt
$$
we have
$$
G(x) = bigllbrack (x-t)f_1(t) bigrrbrack_{t=0}^{t=x} +
int_0^x f_1(t) , dt = f_2(x)
$$
and therefore $G'' = f_2'' = f_1' = f$.
This generalizes to higher anti-derivatives:
$$
int_0^x frac{(x-t)^{n-1}}{(n-1)!} f(t) , dt
= int_0^x int_0^{t_1} cdots int_0^{t_{n-1}} f(t_n) , dt_n cdots dt_1
$$
for $n ge 1$.
answered Dec 6 '18 at 6:53
Martin RMartin R
27.9k33255
$endgroup$
Your proof is correct. An alternative is to use integration by parts. With
$$
f_1(x) = int_0^x f(t) , dt , , quad f_2(x) = int_0^x f_1(t) , dt
$$
we have
$$
G(x) = bigllbrack (x-t)f_1(t) bigrrbrack_{t=0}^{t=x} +
int_0^x f_1(t) , dt = f_2(x)
$$
and therefore $G'' = f_2'' = f_1' = f$.
This generalizes to higher anti-derivatives:
$$
int_0^x frac{(x-t)^{n-1}}{(n-1)!} f(t) , dt
= int_0^x int_0^{t_1} cdots int_0^{t_{n-1}} f(t_n) , dt_n cdots dt_1
$$
for $n ge 1$.
answered Dec 6 '18 at 6:53
Martin RMartin R
27.9k33255
edited Dec 6 '18 at 7:37
answered Dec 6 '18 at 6:53
Martin RMartin R
27.9k33255
answered Dec 6 '18 at 6:53
Martin RMartin R
27.9k33255
answered Dec 6 '18 at 6:53
Martin RMartin R
27.9k33255
27.9k33255
add a comment |
add a comment |
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$begingroup$
Looks fine. You can also use the Leibniz integral rule.
$endgroup$
– Mattos
Dec 6 '18 at 6:15
$begingroup$
@Mattos, thank you I will also try this problem with Leibniz for practise
$endgroup$
– Albert Diaz
Dec 6 '18 at 6:22