Inequality with indefinite integrals and extreme points
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f:[0,1]->R is continuous and nonconstant function. F is the indefinite integral of f such that F(0)=F(1)=0. m and M are the minimum and the maximum values of f.Now we define g:[0,1]->R , g(x)=xf(x) and the indefinite integral of g,G, such that G(0)=0. How can you prove that $|G(1)|<=-mM/(2(M-m))$ ?
calculus functions boundary-value-problem
asked Nov 18 at 19:13
Andrei Gabor
556
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f:[0,1]->R is continuous and nonconstant function. F is the indefinite integral of f such that F(0)=F(1)=0. m and M are the minimum and the maximum values of f.Now we define g:[0,1]->R , g(x)=xf(x) and the indefinite integral of g,G, such that G(0)=0. How can you prove that $|G(1)|<=-mM/(2(M-m))$ ?
calculus functions boundary-value-problem
asked Nov 18 at 19:13
Andrei Gabor
556
Use integration by parts. Perhaps that can help with estimation.
– Anurag A
Nov 18 at 19:17
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
f:[0,1]->R is continuous and nonconstant function. F is the indefinite integral of f such that F(0)=F(1)=0. m and M are the minimum and the maximum values of f.Now we define g:[0,1]->R , g(x)=xf(x) and the indefinite integral of g,G, such that G(0)=0. How can you prove that $|G(1)|<=-mM/(2(M-m))$ ?
calculus functions boundary-value-problem
asked Nov 18 at 19:13
Andrei Gabor
556
f:[0,1]->R is continuous and nonconstant function. F is the indefinite integral of f such that F(0)=F(1)=0. m and M are the minimum and the maximum values of f.Now we define g:[0,1]->R , g(x)=xf(x) and the indefinite integral of g,G, such that G(0)=0. How can you prove that $|G(1)|<=-mM/(2(M-m))$ ?
calculus functions boundary-value-problem
calculus functions boundary-value-problem
asked Nov 18 at 19:13
Andrei Gabor
556
asked Nov 18 at 19:13
Andrei Gabor
556
asked Nov 18 at 19:13
Andrei Gabor
556
asked Nov 18 at 19:13
Andrei Gabor
556
asked Nov 18 at 19:13
Andrei Gabor
556
556
Use integration by parts. Perhaps that can help with estimation.
– Anurag A
Nov 18 at 19:17
add a comment |
Use integration by parts. Perhaps that can help with estimation.
– Anurag A
Nov 18 at 19:17
Use integration by parts. Perhaps that can help with estimation.
– Anurag A
Nov 18 at 19:17
Use integration by parts. Perhaps that can help with estimation.
– Anurag A
Nov 18 at 19:17
add a comment |
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Use integration by parts. Perhaps that can help with estimation.
– Anurag A
Nov 18 at 19:17