Solution Verifivation
$begingroup$
Problem
Let $f(x)$ be a continuous function, and
$$int_0^x f(x-t){rm d}t=e^{-2x}(x+1)-1.$$
Find $int_0^1 f(x){rm d}x$ and $f(x).$
Solution
Differentiate the both sides of the assumption equality with respect to $x$. For
the left hand side, we make a substitution $x-t=:u$ and obtain
begin{align*}
frac{{rm d}}{{rm d}x}int_0^xf(x-t){rm d}t&=frac{{rm d}}{{rm d}(u+t)}int_x^0f(u){rm d}(x-u)\
&=frac{{rm d}}{{rm d}u}int_0^xf(u){rm d}u\
&=f(x).
end{align*}
As for the right hand side, we may directly obtain
$$frac{{rm d}}{{rm d}x}[e^{-2x}(x+1)-1]=-e^{-2x}(2x+1).$$
Thus
$$f(x)=-e^{-2x}(2x+1),$$
$$int_0^1f(x){rm d}x=[e^{-2x}(x+1)]_0^1=frac{2}{e^2}-1.$$
proof-verification definite-integrals
asked Dec 15 '18 at 5:55
mengdie1982mengdie1982
4,882618
$endgroup$
|
show 1 more comment
$begingroup$
Problem
Let $f(x)$ be a continuous function, and
$$int_0^x f(x-t){rm d}t=e^{-2x}(x+1)-1.$$
Find $int_0^1 f(x){rm d}x$ and $f(x).$
Solution
Differentiate the both sides of the assumption equality with respect to $x$. For
the left hand side, we make a substitution $x-t=:u$ and obtain
begin{align*}
frac{{rm d}}{{rm d}x}int_0^xf(x-t){rm d}t&=frac{{rm d}}{{rm d}(u+t)}int_x^0f(u){rm d}(x-u)\
&=frac{{rm d}}{{rm d}u}int_0^xf(u){rm d}u\
&=f(x).
end{align*}
As for the right hand side, we may directly obtain
$$frac{{rm d}}{{rm d}x}[e^{-2x}(x+1)-1]=-e^{-2x}(2x+1).$$
Thus
$$f(x)=-e^{-2x}(2x+1),$$
$$int_0^1f(x){rm d}x=[e^{-2x}(x+1)]_0^1=frac{2}{e^2}-1.$$
proof-verification definite-integrals
asked Dec 15 '18 at 5:55
mengdie1982mengdie1982
4,882618
$endgroup$
$begingroup$
Could you elaborate on this equality? I'm having trouble seeing where it comes from myself. $$frac{{rm d}}{{rm d}(u+t)}int_x^0f(u){rm d}(x-u)=frac{{rm d}}{{rm d}u}int_0^xf(u){rm d}(u)$$
$endgroup$
– Eevee Trainer
Dec 15 '18 at 5:59
$begingroup$
which one? $d(u+t)=du$？
$endgroup$
– mengdie1982
Dec 15 '18 at 6:01
$begingroup$
Never mind, I'm dumb. The multivariate nature of the integral and how you did the $u$-substitution just confused me, never seen it done quite that way.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 6:05
1
$begingroup$
Anyhow, everything seems correct to me, but I'd wait for confirmation from someone else just in case owing to my own slight confusion earlier (i.e. I might not be the best person to evaluate your answer, even if I agree).
$endgroup$
– Eevee Trainer
Dec 15 '18 at 6:10
$begingroup$
@mengdie1982 Correct to me!
$endgroup$
– Nosrati
Dec 15 '18 at 6:50
|
show 1 more comment
$begingroup$
Problem
Let $f(x)$ be a continuous function, and
$$int_0^x f(x-t){rm d}t=e^{-2x}(x+1)-1.$$
Find $int_0^1 f(x){rm d}x$ and $f(x).$
Solution
Differentiate the both sides of the assumption equality with respect to $x$. For
the left hand side, we make a substitution $x-t=:u$ and obtain
begin{align*}
frac{{rm d}}{{rm d}x}int_0^xf(x-t){rm d}t&=frac{{rm d}}{{rm d}(u+t)}int_x^0f(u){rm d}(x-u)\
&=frac{{rm d}}{{rm d}u}int_0^xf(u){rm d}u\
&=f(x).
end{align*}
As for the right hand side, we may directly obtain
$$frac{{rm d}}{{rm d}x}[e^{-2x}(x+1)-1]=-e^{-2x}(2x+1).$$
Thus
$$f(x)=-e^{-2x}(2x+1),$$
$$int_0^1f(x){rm d}x=[e^{-2x}(x+1)]_0^1=frac{2}{e^2}-1.$$
proof-verification definite-integrals
asked Dec 15 '18 at 5:55
mengdie1982mengdie1982
4,882618
$endgroup$
Problem
Let $f(x)$ be a continuous function, and
$$int_0^x f(x-t){rm d}t=e^{-2x}(x+1)-1.$$
Find $int_0^1 f(x){rm d}x$ and $f(x).$
Solution
Differentiate the both sides of the assumption equality with respect to $x$. For
the left hand side, we make a substitution $x-t=:u$ and obtain
begin{align*}
frac{{rm d}}{{rm d}x}int_0^xf(x-t){rm d}t&=frac{{rm d}}{{rm d}(u+t)}int_x^0f(u){rm d}(x-u)\
&=frac{{rm d}}{{rm d}u}int_0^xf(u){rm d}u\
&=f(x).
end{align*}
As for the right hand side, we may directly obtain
$$frac{{rm d}}{{rm d}x}[e^{-2x}(x+1)-1]=-e^{-2x}(2x+1).$$
Thus
$$f(x)=-e^{-2x}(2x+1),$$
$$int_0^1f(x){rm d}x=[e^{-2x}(x+1)]_0^1=frac{2}{e^2}-1.$$
proof-verification definite-integrals
proof-verification definite-integrals
asked Dec 15 '18 at 5:55
mengdie1982mengdie1982
4,882618
asked Dec 15 '18 at 5:55
mengdie1982mengdie1982
4,882618
edited Dec 15 '18 at 5:59
mengdie1982
asked Dec 15 '18 at 5:55
mengdie1982mengdie1982
4,882618
asked Dec 15 '18 at 5:55
mengdie1982mengdie1982
4,882618
asked Dec 15 '18 at 5:55
mengdie1982mengdie1982
4,882618
4,882618
$begingroup$
Could you elaborate on this equality? I'm having trouble seeing where it comes from myself. $$frac{{rm d}}{{rm d}(u+t)}int_x^0f(u){rm d}(x-u)=frac{{rm d}}{{rm d}u}int_0^xf(u){rm d}(u)$$
$endgroup$
– Eevee Trainer
Dec 15 '18 at 5:59
$begingroup$
which one? $d(u+t)=du$？
$endgroup$
– mengdie1982
Dec 15 '18 at 6:01
$begingroup$
Never mind, I'm dumb. The multivariate nature of the integral and how you did the $u$-substitution just confused me, never seen it done quite that way.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 6:05
1
$begingroup$
Anyhow, everything seems correct to me, but I'd wait for confirmation from someone else just in case owing to my own slight confusion earlier (i.e. I might not be the best person to evaluate your answer, even if I agree).
$endgroup$
– Eevee Trainer
Dec 15 '18 at 6:10
$begingroup$
@mengdie1982 Correct to me!
$endgroup$
– Nosrati
Dec 15 '18 at 6:50
|
show 1 more comment
$begingroup$
Could you elaborate on this equality? I'm having trouble seeing where it comes from myself. $$frac{{rm d}}{{rm d}(u+t)}int_x^0f(u){rm d}(x-u)=frac{{rm d}}{{rm d}u}int_0^xf(u){rm d}(u)$$
$endgroup$
– Eevee Trainer
Dec 15 '18 at 5:59
$begingroup$
which one? $d(u+t)=du$？
$endgroup$
– mengdie1982
Dec 15 '18 at 6:01
$begingroup$
Never mind, I'm dumb. The multivariate nature of the integral and how you did the $u$-substitution just confused me, never seen it done quite that way.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 6:05
1
$begingroup$
Anyhow, everything seems correct to me, but I'd wait for confirmation from someone else just in case owing to my own slight confusion earlier (i.e. I might not be the best person to evaluate your answer, even if I agree).
$endgroup$
– Eevee Trainer
Dec 15 '18 at 6:10
$begingroup$
@mengdie1982 Correct to me!
$endgroup$
– Nosrati
Dec 15 '18 at 6:50
$begingroup$
Could you elaborate on this equality? I'm having trouble seeing where it comes from myself. $$frac{{rm d}}{{rm d}(u+t)}int_x^0f(u){rm d}(x-u)=frac{{rm d}}{{rm d}u}int_0^xf(u){rm d}(u)$$
$endgroup$
– Eevee Trainer
Dec 15 '18 at 5:59
$begingroup$
Could you elaborate on this equality? I'm having trouble seeing where it comes from myself. $$frac{{rm d}}{{rm d}(u+t)}int_x^0f(u){rm d}(x-u)=frac{{rm d}}{{rm d}u}int_0^xf(u){rm d}(u)$$
$endgroup$
– Eevee Trainer
Dec 15 '18 at 5:59
$begingroup$
which one? $d(u+t)=du$？
$endgroup$
– mengdie1982
Dec 15 '18 at 6:01
$begingroup$
which one? $d(u+t)=du$？
$endgroup$
– mengdie1982
Dec 15 '18 at 6:01
$begingroup$
Never mind, I'm dumb. The multivariate nature of the integral and how you did the $u$-substitution just confused me, never seen it done quite that way.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 6:05
$begingroup$
Never mind, I'm dumb. The multivariate nature of the integral and how you did the $u$-substitution just confused me, never seen it done quite that way.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 6:05
1
1
$begingroup$
Anyhow, everything seems correct to me, but I'd wait for confirmation from someone else just in case owing to my own slight confusion earlier (i.e. I might not be the best person to evaluate your answer, even if I agree).
$endgroup$
– Eevee Trainer
Dec 15 '18 at 6:10
$begingroup$
Anyhow, everything seems correct to me, but I'd wait for confirmation from someone else just in case owing to my own slight confusion earlier (i.e. I might not be the best person to evaluate your answer, even if I agree).
$endgroup$
– Eevee Trainer
Dec 15 '18 at 6:10
$begingroup$
@mengdie1982 Correct to me!
$endgroup$
– Nosrati
Dec 15 '18 at 6:50
$begingroup$
@mengdie1982 Correct to me!
$endgroup$
– Nosrati
Dec 15 '18 at 6:50
|
show 1 more comment
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$begingroup$
Could you elaborate on this equality? I'm having trouble seeing where it comes from myself. $$frac{{rm d}}{{rm d}(u+t)}int_x^0f(u){rm d}(x-u)=frac{{rm d}}{{rm d}u}int_0^xf(u){rm d}(u)$$
$endgroup$
– Eevee Trainer
Dec 15 '18 at 5:59
$begingroup$
which one? $d(u+t)=du$？
$endgroup$
– mengdie1982
Dec 15 '18 at 6:01
$begingroup$
Never mind, I'm dumb. The multivariate nature of the integral and how you did the $u$-substitution just confused me, never seen it done quite that way.
$endgroup$
– Eevee Trainer
Dec 15 '18 at 6:05
1
$begingroup$
Anyhow, everything seems correct to me, but I'd wait for confirmation from someone else just in case owing to my own slight confusion earlier (i.e. I might not be the best person to evaluate your answer, even if I agree).
$endgroup$
– Eevee Trainer
Dec 15 '18 at 6:10
$begingroup$
@mengdie1982 Correct to me!
$endgroup$
– Nosrati
Dec 15 '18 at 6:50