If-a-b-f-x-dx-a-b-g-x-dx-is-f-x-g-x-true-or-false-

Question Number 131505 by benjo_mathlover last updated on 05/Feb/21

If \int_{a}^{b} f (x) dx = \int_{a}^{b} g (x) dx

is f (x) = g (x) ?

true or false ?

Commented by EDWIN88 last updated on 05/Feb/21

If \underset{a}{\int^{b}} f (x) dx = \underset{a}{\int^{b}} g (x) dx, we does not claim that

f (x) = g (x) . but \underset{a}{\int^{b}} f (x) dx = \underset{a}{\int^{b}} f (a + b - x) dx

then f (x) = f (a + b - x) .

Answered by talminator2856791 last updated on 05/Feb/21

false

Answered by talminator2856791 last updated on 05/Feb/21

take any two straight lines f (x), g (x) of different gradients .

from point of intersection (m; n), take any real p such that

∣ f (p) ∣, ∣ g (p) ∣ ⩽ ∣ n ∣, then

\int_{m - p}^{m + p} f (x) d x = \int_{m - p}^{m + p} g (x) d x

\Rightarrow \int_{a}^{b} f (x) d x = \int_{a}^{b} g (x) d x

the assertion that f (x) = g (x) is false, as f (x) and g (x)

can not be the same line if their gradients are different .

Commented by benjo_mathlover last updated on 05/Feb/21

give me a counterexample

Answered by TheSupreme last updated on 05/Feb/21

c = \frac{\int_{a}^{b} f (x) d x}{b - a}

g (x) = c

f (x) = a n y f u n c t i o n

f (x) \neq c

Commented by talminator2856791 last updated on 05/Feb/21

is this assumption ?

Answered by mr W last updated on 05/Feb/21

Commented by prakash jain last updated on 05/Feb/21

If \int_{a}^{b} f (x) d x = \int_{a}^{b} g (x) d x

for all a, b \in R then f (x) = g (x)

correct ?

Commented by mr W last updated on 05/Feb/21

y e s, i t h i n k .

i f \int_{a}^{b} f (x) d x = \int_{a}^{b} g (x) d x f o r a n y a, b \in R

\Rightarrow \lim_{b \to a} \int_{a}^{b} f (x) d x = \lim_{b \to a} \int_{a}^{b} g (x) d x

\Rightarrow \lim_{b \to a} f (a) (b - a) = \lim_{b \to a} g (a) (b - a)

\Rightarrow f (a) = g (a)

\Rightarrow f (x) = g (x)

Commented by mr W last updated on 05/Feb/21

\int_{a}^{b} f (x) d x = r e d s h a d e d a r e a

\int_{a}^{b} g (x) d x = b l u e s h a d e d a r e a

\int_{a}^{b} f (x) d x = \int_{a}^{b} g (x) d x m e a n s o n l y

t h a t t h e r e d s h a d e d a r e a a n d t h e

b l u e s h a d e d a r e a a r e e q u a l . i t {d o e s n}^{'} t

m e a n t h a t f (x) = g (x)!

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