Integrals: 1/17 ≤ f(x) ≤ 19/68

[nickolas2730]

Use the properties of integrals to show that
1/17smaller or equal to

greater or equal to19/68

where f(x)=1/(1+x⁴) and a=1, b=2
[for the second inequality nd a linear function use the Comparison Theorem]

Here is what i up to

Since f(x) is a decreasing function
thus, when x=2 , f(2) is min, where f(2) = 1/17

i am not sure if i am doing to correct tho...
and i don't know how to use comparison to prove f(x) smaller than 19/68

Sorry i don't really know how to use those symble
thank you

 

[LCKurtz]

Use the properties of integrals to show that
1/17smaller or equal to

greater or equal to19/68

where f(x)=1/(1+x⁴) and a=1, b=2
[for the second inequality nd a linear function use the Comparison Theorem]

Here is what i up to

Since f(x) is a decreasing function
thus, when x=2 , f(2) is min, where f(2) = 1/17

i am not sure if i am doing to correct tho...
and i don't know how to use comparison to prove f(x) smaller than 19/68

Sorry i don't really know how to use those symble
thank you

Ok. So far you have 1/17 ≤ 1/(1+x⁴) so

$$\int_1^2 \frac 1 {17}\, dx \leq \int_1^2 \frac {1}{1+x^4}\, dx$$

so you are half done. Now think about comparing f(x) with the straight line g(x) connecting
(1,f(1)) and (2,f(2)). Can you show f(x) ≤ g(x)? And what do you get?

 

[nickolas2730]

thanks for your reply
but my question is 1+x⁴

i tried (1,f(1)) and (2,f(2)) for 1/(1+x⁴)
and i got an equation y=-(15/34)x + 16/17
and after integrate i got 109/68

am i doing it correct?

 

[LCKurtz]

thanks for your reply
but my question is 1+x⁴

Yes, that's a typo. I fixed it.

i tried (1,f(1)) and (2,f(2)) for 1/(1+x⁴)
and i got an equation y=-(15/34)x + 16/17
and after integrate i got 109/68

am i doing it correct?

The curves form a trapezoid, so you should be able to check the area without calculus. I get a different answer. And remember you have to show that your curve lies under the straight line so you actually have the inequality.

 

[nickolas2730]

i found the area of the straight line g(x)
it is 9/17, right..?
but it is different from the question
and i don't know how to show that g(x)>f(x)
sorry ..

 

[LCKurtz]

i found the area of the straight line g(x)
it is 9/17, right..?
but it is different from the question
and i don't know how to show that g(x)>f(x)
sorry ..

You mean the area under the straight line. And no, it isn't 9/17. You have a trapezoid of height 1 and bases f(1) = 1/2 and and f(2) = 1/17. You don't need calculus to calculate it although both geometry and calculus methods should give you the same answer.

One way to show the curve lies under its secant line on (1,2) would be to show the curve is concave up in that interval.

 

[nickolas2730]

omg,i made a very stupid mistake.
thank you very much
it makes so much sense to me now

thank you