Solution Verifivation












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 5:59







mengdie1982

















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


















  • $begingroup$
    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










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