Evaluate $int_0^frac{pi}{2}f(x){rm d}x$.












4












$begingroup$


Problem



Let $f(x)$ be a continuous function satisfying that $f(x)-cos^2 x=dfrac{1}{pi}displaystyleint_0^{frac{pi}{4}}f(2x){rm d}x$. Evaluate $displaystyleint_0^frac{pi}{2}f(x){rm d}x$.



Solution



Denote $displaystyle int_0^frac{pi}{2}f(x){rm d}x=:I.$ Make a substitution. Let $x=:dfrac{u}{2}$. Then ${rm d}x=dfrac{1}{2}{rm d}u$. It follows that
$$int_0^{frac{pi}{4}}f(2x){rm d}x=frac{1}{2}int_0^{frac{pi}{2}}f(u){rm d}u=frac{I}{2}.$$
Thus, according to the given equality, we have$$f(x)-cos^2 x=frac{I}{2pi}$$
Integrate the both sides from $0$ to $dfrac{pi}{2}$. We obtain $$int_0^frac{pi}{2}f(x){rm d}x-int_0^frac{pi}{2}cos^2 x{rm d}x=int_0^frac{pi}{2}frac{I}{2pi}{rm d}x,$$which implies
$$I-frac{pi}{4}=frac{I}{2pi}cdot left(frac{pi}{2}-0right).$$
As a result,
$$I=frac{pi}{3}.$$






proof-verification definite-integrals







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$endgroup$








  • 2




    $begingroup$
    I think it should be $f(x)-cos^2 x=frac{I}{2pi}$.
    $endgroup$
    – Thomas Shelby
    Jan 12 at 11:27






  • 1




    $begingroup$
    @ThomasShelby Thanks! Corrected!
    $endgroup$
    – mengdie1982
    Jan 12 at 11:31
















4












$begingroup$


Problem



Let $f(x)$ be a continuous function satisfying that $f(x)-cos^2 x=dfrac{1}{pi}displaystyleint_0^{frac{pi}{4}}f(2x){rm d}x$. Evaluate $displaystyleint_0^frac{pi}{2}f(x){rm d}x$.



Solution



Denote $displaystyle int_0^frac{pi}{2}f(x){rm d}x=:I.$ Make a substitution. Let $x=:dfrac{u}{2}$. Then ${rm d}x=dfrac{1}{2}{rm d}u$. It follows that
$$int_0^{frac{pi}{4}}f(2x){rm d}x=frac{1}{2}int_0^{frac{pi}{2}}f(u){rm d}u=frac{I}{2}.$$
Thus, according to the given equality, we have$$f(x)-cos^2 x=frac{I}{2pi}$$
Integrate the both sides from $0$ to $dfrac{pi}{2}$. We obtain $$int_0^frac{pi}{2}f(x){rm d}x-int_0^frac{pi}{2}cos^2 x{rm d}x=int_0^frac{pi}{2}frac{I}{2pi}{rm d}x,$$which implies
$$I-frac{pi}{4}=frac{I}{2pi}cdot left(frac{pi}{2}-0right).$$
As a result,
$$I=frac{pi}{3}.$$






proof-verification definite-integrals







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    I think it should be $f(x)-cos^2 x=frac{I}{2pi}$.
    $endgroup$
    – Thomas Shelby
    Jan 12 at 11:27






  • 1




    $begingroup$
    @ThomasShelby Thanks! Corrected!
    $endgroup$
    – mengdie1982
    Jan 12 at 11:31














4












4








4


1



$begingroup$


Problem



Let $f(x)$ be a continuous function satisfying that $f(x)-cos^2 x=dfrac{1}{pi}displaystyleint_0^{frac{pi}{4}}f(2x){rm d}x$. Evaluate $displaystyleint_0^frac{pi}{2}f(x){rm d}x$.



Solution



Denote $displaystyle int_0^frac{pi}{2}f(x){rm d}x=:I.$ Make a substitution. Let $x=:dfrac{u}{2}$. Then ${rm d}x=dfrac{1}{2}{rm d}u$. It follows that
$$int_0^{frac{pi}{4}}f(2x){rm d}x=frac{1}{2}int_0^{frac{pi}{2}}f(u){rm d}u=frac{I}{2}.$$
Thus, according to the given equality, we have$$f(x)-cos^2 x=frac{I}{2pi}$$
Integrate the both sides from $0$ to $dfrac{pi}{2}$. We obtain $$int_0^frac{pi}{2}f(x){rm d}x-int_0^frac{pi}{2}cos^2 x{rm d}x=int_0^frac{pi}{2}frac{I}{2pi}{rm d}x,$$which implies
$$I-frac{pi}{4}=frac{I}{2pi}cdot left(frac{pi}{2}-0right).$$
As a result,
$$I=frac{pi}{3}.$$






proof-verification definite-integrals







share|cite|improve this question











$endgroup$




Problem



Let $f(x)$ be a continuous function satisfying that $f(x)-cos^2 x=dfrac{1}{pi}displaystyleint_0^{frac{pi}{4}}f(2x){rm d}x$. Evaluate $displaystyleint_0^frac{pi}{2}f(x){rm d}x$.



Solution



Denote $displaystyle int_0^frac{pi}{2}f(x){rm d}x=:I.$ Make a substitution. Let $x=:dfrac{u}{2}$. Then ${rm d}x=dfrac{1}{2}{rm d}u$. It follows that
$$int_0^{frac{pi}{4}}f(2x){rm d}x=frac{1}{2}int_0^{frac{pi}{2}}f(u){rm d}u=frac{I}{2}.$$
Thus, according to the given equality, we have$$f(x)-cos^2 x=frac{I}{2pi}$$
Integrate the both sides from $0$ to $dfrac{pi}{2}$. We obtain $$int_0^frac{pi}{2}f(x){rm d}x-int_0^frac{pi}{2}cos^2 x{rm d}x=int_0^frac{pi}{2}frac{I}{2pi}{rm d}x,$$which implies
$$I-frac{pi}{4}=frac{I}{2pi}cdot left(frac{pi}{2}-0right).$$
As a result,
$$I=frac{pi}{3}.$$






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 Jan 12 at 11:30







mengdie1982

















asked Jan 12 at 11:22









mengdie1982mengdie1982

4,932618




4,932618








  • 2




    $begingroup$
    I think it should be $f(x)-cos^2 x=frac{I}{2pi}$.
    $endgroup$
    – Thomas Shelby
    Jan 12 at 11:27






  • 1




    $begingroup$
    @ThomasShelby Thanks! Corrected!
    $endgroup$
    – mengdie1982
    Jan 12 at 11:31














  • 2




    $begingroup$
    I think it should be $f(x)-cos^2 x=frac{I}{2pi}$.
    $endgroup$
    – Thomas Shelby
    Jan 12 at 11:27






  • 1




    $begingroup$
    @ThomasShelby Thanks! Corrected!
    $endgroup$
    – mengdie1982
    Jan 12 at 11:31








2




2




$begingroup$
I think it should be $f(x)-cos^2 x=frac{I}{2pi}$.
$endgroup$
– Thomas Shelby
Jan 12 at 11:27




$begingroup$
I think it should be $f(x)-cos^2 x=frac{I}{2pi}$.
$endgroup$
– Thomas Shelby
Jan 12 at 11:27




1




1




$begingroup$
@ThomasShelby Thanks! Corrected!
$endgroup$
– mengdie1982
Jan 12 at 11:31




$begingroup$
@ThomasShelby Thanks! Corrected!
$endgroup$
– mengdie1982
Jan 12 at 11:31










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