Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

find the maximum value of f(x) =integration of 0 to x (2t-t²+5)dt

find the maximum value of f(x) =integration of 0 to x (2t-t²+5)dt

Grade:12

3 Answers

lokesh palingi
44 Points
7 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)
MuraliKrishna Medavaram
askIITians Faculty 33 Points
7 years ago

175-1492_Untitled1.jpg
Thanks and Regards,
M.MURALIKRISHNA
askIITians faculty
MuraliKrishna Medavaram
askIITians Faculty 33 Points
7 years ago
a

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free

10th iCAT Scholarship Test Registration Form