Given-f-x-0-1-f-x-dx-x-2-0-2-f-x-dx-x-0-3-f-x-dx-1-then-the-value-of-f-4-

Question Number 123386 by bemath last updated on 25/Nov/20

G i v e n

f (x) = (\underset{0}{\int^{1}} f (x) d x) x^{2} + (\underset{0}{\int^{2}} f (x) d x) x + (\underset{0}{\int^{3}} f (x) d x) + 1

t h e n t h e v a l u e o f f (4) = \dots

Answered by TANMAY PANACEA last updated on 25/Nov/20

f (x) = {a x}^{2} + b x + c + 1

a = \int_{0}^{1} ({a x}^{2} + b x + c + 1) d x

a =∣ \frac{{a x}^{3}}{3} + \frac{{b x}^{2}}{2} + c x + x ∣_{0}^{1}

a = \frac{a}{3} + \frac{b}{2} + c + 1 \dots . . (1 s t e q n)

b =∣ \frac{{a x}^{3}}{3} + \frac{{b x}^{2}}{2} + c x + x ∣_{0}^{2}

b = \frac{8 a}{3} + \frac{4 b}{2} + 2 c + 2 \dots . + (2 n d e q n)

c =∣ \frac{{a x}^{3}}{3} + \frac{{b x}^{2}}{2} + c x + x ∣_{0}^{3}

c = \frac{27 a}{3} + \frac{9 b}{4} + 3 c + 3 \dots . (3 r d e q n)

w e h a v e t o s o l v e t o f i n d (a, b a n d c)

f (x) = {a x}^{2} + b x + c + 1

n x t t o p u t x = 4

p l s t r y f r o m h e r e . .