Find $int_{0}^{1}f(x)g(x)dx$Two simple statements about continuous and monotonic functionsExchanging limits...
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Find $int_{0}^{1}f(x)g(x)dx$
Two simple statements about continuous and monotonic functionsExchanging limits and Riemann IntegralEvaluate :$displaystylelim_{hto 0}frac 1 hint_{-infty}^infty gleft(frac x hright)f(x)mathbb dx$$int_{1}^{infty} h(x) dx$ converges $Rightarrow$ $h$ is bounded in $[1, infty)$Evaluate $lim_{epsilonrightarrow0}frac{1}{epsilon}int_{-infty}^{infty}rholeft(frac{x}{epsilon}right)f(x)dx$Prove that the integral $int_{0}^{infty}bigg|frac{sin(pi x)}{x}bigg|dx$ divergesThe value of $a$ besides $1$ for which $gamma =lim_{nrightarrowinfty}left(sum_{k=1}^{n}frac{1}{k^a}-int_{1}^{n}frac{1}{x^a}dxright)$For a continuous function with $int_{0}^{1} f(x)dx=1$, and $M=max{f(x)}$, show that $1-frac{1}{2M}geqint_{0}^{1}xf(x)dxgeqfrac{1}{2M}$Let $f:mathbb{R}rightarrowmathbb{R}$ be a continuous function such that $int_{x}^{1}f(t)dtgeq(1-x)^{2}$. Prove that $f(1)=0$If $2int_{0}^{1} xf(x)dxgeqint_{0}^{1}(f(x))^{2}dx$, prove than $int_{0}^{1}(f(x))^2dxgeqfrac{4}{3}$
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Let $f:[0,1]rightarrowmathbb{R}$ a continuous and positive number, such that there is a real number $c$ such that $int_{0}^{1} f(x)x^{k}dx=c^k$, for $k={0,1,...,n}$ , (1). For a function $g:[0,1]rightarrowmathbb{R}$, continuous, find $int_{0}^{1}f(x)g(x)dx$ and show that $cin[0,1]$. Because the integrand (1) is bounded below by $0$, $cgeq0$ and by the mean value theorem, there is an $ain[0,1]$ such that $int_{0}^{1} f(x)x^{k}dx=frac{f(a)}{k+1}$, and from here by the fact that for $lim_{krightarrowinfty} frac{f(a)}{k+1}=0$, $lim_{krightarrowinfty} c^k=0$ and so $c$ can't be bigger than $1$. However, I am unable to use this in proving the second task.
calculus integration definite-integrals
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$begingroup$
Let $f:[0,1]rightarrowmathbb{R}$ a continuous and positive number, such that there is a real number $c$ such that $int_{0}^{1} f(x)x^{k}dx=c^k$, for $k={0,1,...,n}$ , (1). For a function $g:[0,1]rightarrowmathbb{R}$, continuous, find $int_{0}^{1}f(x)g(x)dx$ and show that $cin[0,1]$. Because the integrand (1) is bounded below by $0$, $cgeq0$ and by the mean value theorem, there is an $ain[0,1]$ such that $int_{0}^{1} f(x)x^{k}dx=frac{f(a)}{k+1}$, and from here by the fact that for $lim_{krightarrowinfty} frac{f(a)}{k+1}=0$, $lim_{krightarrowinfty} c^k=0$ and so $c$ can't be bigger than $1$. However, I am unable to use this in proving the second task.
calculus integration definite-integrals
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add a comment |
$begingroup$
Let $f:[0,1]rightarrowmathbb{R}$ a continuous and positive number, such that there is a real number $c$ such that $int_{0}^{1} f(x)x^{k}dx=c^k$, for $k={0,1,...,n}$ , (1). For a function $g:[0,1]rightarrowmathbb{R}$, continuous, find $int_{0}^{1}f(x)g(x)dx$ and show that $cin[0,1]$. Because the integrand (1) is bounded below by $0$, $cgeq0$ and by the mean value theorem, there is an $ain[0,1]$ such that $int_{0}^{1} f(x)x^{k}dx=frac{f(a)}{k+1}$, and from here by the fact that for $lim_{krightarrowinfty} frac{f(a)}{k+1}=0$, $lim_{krightarrowinfty} c^k=0$ and so $c$ can't be bigger than $1$. However, I am unable to use this in proving the second task.
calculus integration definite-integrals
$endgroup$
Let $f:[0,1]rightarrowmathbb{R}$ a continuous and positive number, such that there is a real number $c$ such that $int_{0}^{1} f(x)x^{k}dx=c^k$, for $k={0,1,...,n}$ , (1). For a function $g:[0,1]rightarrowmathbb{R}$, continuous, find $int_{0}^{1}f(x)g(x)dx$ and show that $cin[0,1]$. Because the integrand (1) is bounded below by $0$, $cgeq0$ and by the mean value theorem, there is an $ain[0,1]$ such that $int_{0}^{1} f(x)x^{k}dx=frac{f(a)}{k+1}$, and from here by the fact that for $lim_{krightarrowinfty} frac{f(a)}{k+1}=0$, $lim_{krightarrowinfty} c^k=0$ and so $c$ can't be bigger than $1$. However, I am unable to use this in proving the second task.
calculus integration definite-integrals
calculus integration definite-integrals
asked Mar 15 at 8:52
user651754
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I believe the hypothesis should be for all $k neq 0$. If it is for $0leq k leq n$ for some fixed $n$ it is not possible to determine $int_0^{1}f(x)g(x), dx$.
For any polynomial $p$ we get $int_0^{1}f(x)p(x), dx=p(c)$ by just taking linear combinations. By Weierstrass theorem $g$ can be approximated uniformly by polynomials. This gives $int_0^{1}f(x)g(x), dx=g(c)$.
answered Mar 15 at 8:59
Kavi Rama MurthyKavi Rama Murthy
70.1k53170
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$begingroup$
I believe the hypothesis should be for all $k neq 0$. If it is for $0leq k leq n$ for some fixed $n$ it is not possible to determine $int_0^{1}f(x)g(x), dx$.
For any polynomial $p$ we get $int_0^{1}f(x)p(x), dx=p(c)$ by just taking linear combinations. By Weierstrass theorem $g$ can be approximated uniformly by polynomials. This gives $int_0^{1}f(x)g(x), dx=g(c)$.
answered Mar 15 at 8:59
Kavi Rama MurthyKavi Rama Murthy
70.1k53170
$endgroup$
add a comment |
$begingroup$
I believe the hypothesis should be for all $k neq 0$. If it is for $0leq k leq n$ for some fixed $n$ it is not possible to determine $int_0^{1}f(x)g(x), dx$.
For any polynomial $p$ we get $int_0^{1}f(x)p(x), dx=p(c)$ by just taking linear combinations. By Weierstrass theorem $g$ can be approximated uniformly by polynomials. This gives $int_0^{1}f(x)g(x), dx=g(c)$.
answered Mar 15 at 8:59
Kavi Rama MurthyKavi Rama Murthy
70.1k53170
$endgroup$
add a comment |
$begingroup$
I believe the hypothesis should be for all $k neq 0$. If it is for $0leq k leq n$ for some fixed $n$ it is not possible to determine $int_0^{1}f(x)g(x), dx$.
For any polynomial $p$ we get $int_0^{1}f(x)p(x), dx=p(c)$ by just taking linear combinations. By Weierstrass theorem $g$ can be approximated uniformly by polynomials. This gives $int_0^{1}f(x)g(x), dx=g(c)$.
answered Mar 15 at 8:59
Kavi Rama MurthyKavi Rama Murthy
70.1k53170
$endgroup$
I believe the hypothesis should be for all $k neq 0$. If it is for $0leq k leq n$ for some fixed $n$ it is not possible to determine $int_0^{1}f(x)g(x), dx$.
For any polynomial $p$ we get $int_0^{1}f(x)p(x), dx=p(c)$ by just taking linear combinations. By Weierstrass theorem $g$ can be approximated uniformly by polynomials. This gives $int_0^{1}f(x)g(x), dx=g(c)$.
answered Mar 15 at 8:59
Kavi Rama MurthyKavi Rama Murthy
70.1k53170
edited Mar 15 at 9:14
answered Mar 15 at 8:59
Kavi Rama MurthyKavi Rama Murthy
70.1k53170
answered Mar 15 at 8:59
Kavi Rama MurthyKavi Rama Murthy
70.1k53170
answered Mar 15 at 8:59
Kavi Rama MurthyKavi Rama Murthy
70.1k53170
70.1k53170
add a comment |
add a comment |
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