"Let a function ƒ : [0, 5] $\to$ R be continuous, ƒ(1) = 3 and F be defined as : $F(x) = \int\limits_1^x {{t^2}g(t)dt}$ , where $g(t) = \int\limits_1^t {f(u)du}$ Then for the function F, the point x = 1 is :"

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Accepted Answer

"$F(x) = \int\limits_1^x {{t^2}g(t)dt}$

$\Rightarrow$ F'(x) = x²g(x) = x²$\int\limits_1^t {f(u)du}$

$\therefore$ F'(1) = (1)(0) = 0

Now, F''(x) = 2xg(x) + x²g'(x)

F''(1) = 2g(1) + g'(1) = 0 + g'(1) = 3

[ As g'(t) = f(t); g'(1) = f'(1) = 3 ]

So, at x = 1, F'(1) = 0 and F"(1) = 3 > 0

$\therefore$ For the function f(x), x = 1 is a point of local minima. "

Let a function ƒ : [0, 5] $\to$ R be continuous, ƒ(1) = 3 and F be defined as :

$F(x) = \int\limits_1^x {{t^2}g(t)dt}$ , where $g(t) = \int\limits_1^t {f(u)du}$

Then for the function F, the point x = 1 is :

Solution

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Let a function ƒ : [0, 5] $\to$ R be continuous, ƒ(1) = 3 and F be defined as : $F(x) = \int\limits_1^x {{t^2}g(t)dt}$ , where $g(t) = \int\limits_1^t {f(u)du}$ Then for the function F, the point x = 1 is :

Solution

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More Definite Integration questions

Drill 25 more like these. Every day. Free.

Let a function ƒ : [0, 5] $\to$ R be continuous, ƒ(1) = 3 and F be defined as :

$F(x) = \int\limits_1^x {{t^2}g(t)dt}$ , where $g(t) = \int\limits_1^t {f(u)du}$

Then for the function F, the point x = 1 is :