Let a function f :[0, 5] \toR be continuous f(1) = 3 and F be defined as : F(x)=\int^{x}_{1}t^2g(t)dt, where g(t)=\int^{t}_{1}f(u)du Then for the function F, the point x = 1 is : [9 Jan 2020 Shift 2]

Correct Answer is : a point of local minima.

Definite Integrals Part 1

Section: Mathematics

Question : 23 of 100

Marks: +1, -0

Let a function f:[0,5]→R be continuous f(1) = 3 and F be defined as :
F(x)=

∫

t2g(t)dt, where g(t)=

∫

f(u)du
Then for the function F, the point x = 1 is :

[9 Jan 2020 Shift 2]

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