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)