Integral Calculus> find-maximum-value-of-f-x...

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

Pradeeep Anand , 11 Years ago

Grade 11

3 Answers

Parvez ali

Last Activity: 11 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)₀^1+sqrt(6)

=17/3 + 4sqrt(6)

Thanks & Regards

Parvez Ali,

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Approved

Pradeeep Anand

Last Activity: 11 Years ago

thanks a lot

lokesh palingi

Last Activity: 11 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)