If f is continuous function and Fx- -0x 2t-3-t2f udu dt- then the value of - F-2-f2 -

∴ F(x)= ∫_0^x ( (2t+3)∫_t^2f (u)du )dt Differentiating both sides w.r.t. x, then F'(x)=(2x+3).∫_x^2f(u)du Again differentiating both sides w.r.t. x, then F”( ...

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Standard XII

Mathematics

Differentiation under Integral Sign

If f is conti...

Question

If $f$ is continuous function and $F (x) = \int_{0}^{x} ((2 t + 3) \int_{t}^{2} f (u) d u) d t$ , then the value of $∣ ∣ ∣ \frac{F^{''} (2)}{f (2)} ∣ ∣ ∣$

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Solution

The correct option is C 7
$∴ F (x) = \int_{0}^{x} ((2 t + 3) \int_{t}^{2} f (u) d u) d t$
Differentiating both sides w.r.t. x, then
$F^{'} (x) = (2 x + 3) . \int_{x}^{2} f (u) d u$
Again differentiating both sides w.r.t. x, then
$F^{''} (x) = (2 x + 3) . (0 - f (x)) + \int_{x}^{2} f (u) d u \times 2 ∴ F^{''} (2) = - 7 f (2) + 0 \Rightarrow ∣ ∣ ∣ \frac{F^{''} (2)}{f (2)} ∣ ∣ ∣ = 7$

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If f is continuous function and F(x)=∫x0((2t+3)∫2tf(u)du)dt, then the value of ∣∣∣F′′(2)f(2)∣∣∣

The correct option is C 7∴F(x)=∫x0((2t+3)∫2tf(u)du)dt Differentiating both sides w.r.t. x, then F′(x)=(2x+3).∫2xf(u)du Again differentiating both sides w.r.t. x, then F′′(x)=(2x+3).(0−f(x))+∫2xf(u)du×2∴F′′(2)=−7f(2)+0⇒∣∣∣F′′(2)f(2)∣∣∣=7

If $f$ is continuous function and $F (x) = \int_{0}^{x} ((2 t + 3) \int_{t}^{2} f (u) d u) d t$ , then the value of $∣ ∣ ∣ \frac{F^{''} (2)}{f (2)} ∣ ∣ ∣$