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find maximum value of f(x) where f(x)=integral 0 to x (2t−t²+5)dt

find maximum value of f(x) where f(x)=integral 0 to x (2t−t²+5)dt

Grade:11

3 Answers

Parvez ali
askIITians Faculty 47 Points
10 years ago
on integrating using lebnitz theorem for differantiation under integral sign
df(x)/dx=2x-x2+5
for maxima and minima df(x)/dx=0 =>2x-x2+5=0 => x= 1+sqrt(6),1-sqrt(6)
d2f(x)/dx2=2-2x ,d2f(x)/dx2=sqrt(6) for x= 1-sqrt(6) and ,d2f(x)/dx2=-sqrt(6) for x= 1+sqrt(6)
thus f(x) is maximum for x= 1+sqrt(6) as,d2f(x)/dx2=-sqrt(6)<0
thus maximum value of f(x)=integral 0 to1+sqrt(6)(2t-t²+5)dt
= (t2-t3/3+5t)01+sqrt(6)
=17/3 + 4sqrt(6)

Thanks & Regards
Parvez Ali,
askIITians Faculty

Pradeeep Anand
40 Points
10 years ago
thanks a lot
lokesh palingi
44 Points
10 years ago
on integrating using lebnitz theorem for differantiation under integral sign df(x)/dx=2x-x2+5 for maxima and minima df(x)/dx=0 =>2x-x2+5=0 => x= 1+sqrt(6),1-sqrt(6) d2f(x)/dx2=2-2x ,d2f(x)/dx2=sqrt(6) for x= 1-sqrt(6) and ,d2f(x)/dx2=-sqrt(6) for x= 1+sqrt(6) thus f(x) is maximum for x= 1+sqrt(6) as,d2f(x)/dx2=-sqrt(6)<0 thus maximum value of f(x)=integral 0 to1+sqrt(6)(2t-t²+5)dt = (t2-t3/3+5t)01+sqrt(6) =17/3 + 4sqrt(6)

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