Math Question and Answers


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 59846 by bhanukumarb2@gmail.com last updated on 15/May/19

Commented by MJS last updated on 15/May/19

$do you have an answer ?$ $it seems to be \frac{1}{4}$
Commented by bhanukumarb2@gmail.com last updated on 16/May/19

$t h a n k s s i r$
Commented by bhanukumarb2@gmail.com last updated on 16/May/19

$r i g h t$
Commented by bhanukumarb2@gmail.com last updated on 16/May/19

$u r$
Commented by bhanukumarb2@gmail.com last updated on 16/May/19

$u r a p p r o c h$
Commented by MJS last updated on 16/May/19
$f(x)= { ((1; 0≤x≤a)),((−c; a≤x≤1)) :} I_1 =∫_0 ^a 1dx+∫_a ^1 −cdx=0 ⇒ c=(a/(1−a)) I_2 =∫_0 ^a 1^3 dx+∫_a ^1 ((a/(1−a)))^3 dx=((a(1−2a))/((a−1)^2 )) (d/da)[((a(1−2a))/((a−1)^2 ))]=0 ((3a−1)/((a−1)^3 ))=0 ⇒ a=(1/3) ⇒ c=(1/2) ⇒ I_2 =(1/4) the idea is, make f(x)=1 in the longest possible interval and keep it constant and <1 within the rest, because 1^3 =1 and c^3 <c$
$f (x) = {\begin{cases} 1; 0 ⩽ x ⩽ a \\ - c; a ⩽ x ⩽ 1 \end{cases}$ $I_{1} = \underset{0}{\int^{a}} 1 d x + \underset{a}{\int^{1}} - c d x = 0 \Rightarrow c = \frac{a}{1 - a}$ $I_{2} = \underset{0}{\int^{a}} 1^{3} d x + \underset{a}{\int^{1}} {(\frac{a}{1 - a})}^{3} d x = \frac{a (1 - 2 a)}{{(a - 1)}^{2}}$ $\frac{d}{d a} [\frac{a (1 - 2 a)}{{(a - 1)}^{2}}] = 0$ $\frac{3 a - 1}{{(a - 1)}^{3}} = 0 \Rightarrow a = \frac{1}{3} \Rightarrow c = \frac{1}{2} \Rightarrow I_{2} = \frac{1}{4}$ $the idea is, make f (x) = 1 in the longest$ $possible interval and keep it constant and$ $< 1 within the rest, because 1^{3} = 1 and c^{3} < c$
Commented by bhanukumarb2@gmail.com last updated on 16/May/19

$h o w u g e t s u c h t h o u g h t . h o w i c a n t h i n k$ $l i k e t h i s$
Commented by bhanukumarb2@gmail.com last updated on 16/May/19

$c s h o u l d b e l i k e c = a / . . . i h v e s e e n t h i s a p p r o c h$ $i n m s e t o . . s a m e d o u b t t h a t t i m e . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum