Using comparison theorem for integrals to prove an inequality
The problem states to show that $int_{0}^{frac{pi}{2}} frac{sin(x)}{x(x+5)} dx < frac{pi}{10}$ using the fact that $int_{a}^{b}f(x)dx<int_{a}^{b}g(x)dx$ and $sin(x)<x$.
I attempt to set up the inequality as such, but I am having a hard time finding a continuous g(x) on the bounds that is easily integrable so that the inequality holds. Maybe I can find some $c in [a,b]$ s.t. $g(c) > f(c)$ which would give me a constant that is easily integrable, but I am not sure if this will prove that $int_{0}^{frac{pi}{2}} frac{sin(x)}{x(x+5)} dx < frac{pi}{10}$. If someone who is more creative than me could help me out that would be awesome.
real-analysis integration
asked Nov 29 '18 at 3:31
hkj447
244
add a comment |
The problem states to show that $int_{0}^{frac{pi}{2}} frac{sin(x)}{x(x+5)} dx < frac{pi}{10}$ using the fact that $int_{a}^{b}f(x)dx<int_{a}^{b}g(x)dx$ and $sin(x)<x$.
I attempt to set up the inequality as such, but I am having a hard time finding a continuous g(x) on the bounds that is easily integrable so that the inequality holds. Maybe I can find some $c in [a,b]$ s.t. $g(c) > f(c)$ which would give me a constant that is easily integrable, but I am not sure if this will prove that $int_{0}^{frac{pi}{2}} frac{sin(x)}{x(x+5)} dx < frac{pi}{10}$. If someone who is more creative than me could help me out that would be awesome.
real-analysis integration
asked Nov 29 '18 at 3:31
hkj447
244
add a comment |
The problem states to show that $int_{0}^{frac{pi}{2}} frac{sin(x)}{x(x+5)} dx < frac{pi}{10}$ using the fact that $int_{a}^{b}f(x)dx<int_{a}^{b}g(x)dx$ and $sin(x)<x$.
I attempt to set up the inequality as such, but I am having a hard time finding a continuous g(x) on the bounds that is easily integrable so that the inequality holds. Maybe I can find some $c in [a,b]$ s.t. $g(c) > f(c)$ which would give me a constant that is easily integrable, but I am not sure if this will prove that $int_{0}^{frac{pi}{2}} frac{sin(x)}{x(x+5)} dx < frac{pi}{10}$. If someone who is more creative than me could help me out that would be awesome.
real-analysis integration
asked Nov 29 '18 at 3:31
hkj447
244
The problem states to show that $int_{0}^{frac{pi}{2}} frac{sin(x)}{x(x+5)} dx < frac{pi}{10}$ using the fact that $int_{a}^{b}f(x)dx<int_{a}^{b}g(x)dx$ and $sin(x)<x$.
I attempt to set up the inequality as such, but I am having a hard time finding a continuous g(x) on the bounds that is easily integrable so that the inequality holds. Maybe I can find some $c in [a,b]$ s.t. $g(c) > f(c)$ which would give me a constant that is easily integrable, but I am not sure if this will prove that $int_{0}^{frac{pi}{2}} frac{sin(x)}{x(x+5)} dx < frac{pi}{10}$. If someone who is more creative than me could help me out that would be awesome.
real-analysis integration
real-analysis integration
asked Nov 29 '18 at 3:31
hkj447
244
asked Nov 29 '18 at 3:31
hkj447
244
asked Nov 29 '18 at 3:31
hkj447
244
asked Nov 29 '18 at 3:31
hkj447
244
asked Nov 29 '18 at 3:31
hkj447
244
244
add a comment |
add a comment |
1 Answer
1
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oldest
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HINTS:
Note that for $xin [0,pi/2]$, we have
$$0le frac{sin(x)}{x}le 1$$
and
$$0le frac{1}{x+5}le frac15$$
answered Nov 29 '18 at 3:35
Mark Viola
130k1274170
ah thank you, I was trying to put the x on top and getting nasty integrals.
– hkj447
Nov 29 '18 at 3:39
You're welcome. My pleasure.
– Mark Viola
Nov 29 '18 at 3:43
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
HINTS:
Note that for $xin [0,pi/2]$, we have
$$0le frac{sin(x)}{x}le 1$$
and
$$0le frac{1}{x+5}le frac15$$
answered Nov 29 '18 at 3:35
Mark Viola
130k1274170
ah thank you, I was trying to put the x on top and getting nasty integrals.
– hkj447
Nov 29 '18 at 3:39
You're welcome. My pleasure.
– Mark Viola
Nov 29 '18 at 3:43
add a comment |
HINTS:
Note that for $xin [0,pi/2]$, we have
$$0le frac{sin(x)}{x}le 1$$
and
$$0le frac{1}{x+5}le frac15$$
answered Nov 29 '18 at 3:35
Mark Viola
130k1274170
ah thank you, I was trying to put the x on top and getting nasty integrals.
– hkj447
Nov 29 '18 at 3:39
You're welcome. My pleasure.
– Mark Viola
Nov 29 '18 at 3:43
add a comment |
HINTS:
Note that for $xin [0,pi/2]$, we have
$$0le frac{sin(x)}{x}le 1$$
and
$$0le frac{1}{x+5}le frac15$$
answered Nov 29 '18 at 3:35
Mark Viola
130k1274170
HINTS:
Note that for $xin [0,pi/2]$, we have
$$0le frac{sin(x)}{x}le 1$$
and
$$0le frac{1}{x+5}le frac15$$
answered Nov 29 '18 at 3:35
Mark Viola
130k1274170
answered Nov 29 '18 at 3:35
Mark Viola
130k1274170
answered Nov 29 '18 at 3:35
Mark Viola
130k1274170
answered Nov 29 '18 at 3:35
Mark Viola
130k1274170
130k1274170
ah thank you, I was trying to put the x on top and getting nasty integrals.
– hkj447
Nov 29 '18 at 3:39
You're welcome. My pleasure.
– Mark Viola
Nov 29 '18 at 3:43
add a comment |
ah thank you, I was trying to put the x on top and getting nasty integrals.
– hkj447
Nov 29 '18 at 3:39
You're welcome. My pleasure.
– Mark Viola
Nov 29 '18 at 3:43
ah thank you, I was trying to put the x on top and getting nasty integrals.
– hkj447
Nov 29 '18 at 3:39
ah thank you, I was trying to put the x on top and getting nasty integrals.
– hkj447
Nov 29 '18 at 3:39
You're welcome. My pleasure.
– Mark Viola
Nov 29 '18 at 3:43
You're welcome. My pleasure.
– Mark Viola
Nov 29 '18 at 3:43
add a comment |
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