Is-definite-integral-can-have-negative-value-Because-I-think-a-b-f-x-dx-is-total-area-below-graph-f-x-from-x-a-until-x-b-so-it-can-t-be-negative-

Question Number 20202 by Joel577 last updated on 24/Aug/17

Is definite integral can have negative value ?

Because I think \int_{a}^{b} f (x) d x is total area below

graph f (x) from x = a until x = b, so it {can}^{'} t

be negative

Commented by ajfour last updated on 24/Aug/17

c a n n o t f (x) i t s e l f b e n e g a t i v e f o r

a ⩽ x ⩽ b ?

S o \int_{a}^{b} f (x) d x c a n b e n e g a t i v e,

p o s i t i v e o r z e r o, d e p e n d i n g o n

v a l u e s o f f (x) i n t h e l e n g t h x = a

t o x = b .

Commented by Joel577 last updated on 24/Aug/17

In example I = \int_{0}^{π / 2} x \cos 2 x d x

the result is - \frac{1}{2} . Is it true or not ?

Commented by ajfour last updated on 24/Aug/17

\int_{π}^{2 π} \sin x d x = ?, \int_{- 1}^{0} x d x = ? .

\int x \cos 2 x d x = \frac{x \sin 2 x}{2} ∣_{0}^{π / 2} - \frac{1}{2} \int_{0}^{π / 2} \sin 2 x d x

= 0 + \frac{\cos 2 x}{4} ∣_{0}^{π / 2} = \frac{- 1 - 1}{4} = - \frac{1}{2} .

Commented by Joel577 last updated on 24/Aug/17

But if I separate it into

\int_{0}^{π / 4} x . \cos 2 x d x + \int_{π / 4}^{π / 2} x . \cos 2 x d x

it {isn}^{'} t - \frac{1}{2}

Commented by ajfour last updated on 24/Aug/17

[\frac{x \sin 2 x}{2} + \frac{\cos 2 x}{4}] ∣_{0}^{π / 4} + [\frac{x \sin 2 x}{2} + \frac{\cos 2 x}{4}] ∣_{π / 4}^{π / 2}

= (\frac{π}{8} - \frac{1}{4}) + (- \frac{1}{4} - \frac{π}{8}) = - \frac{1}{2} .

Commented by Joel577 last updated on 24/Aug/17

Commented by Joel577 last updated on 24/Aug/17

I have a little doubt about this

The area between \frac{π}{4} and \frac{π}{2} is below x - axis

So, the integral must be positive, {isn}^{'} t it ?

Commented by Joel577 last updated on 24/Aug/17

Please explain with more detail Sir Ajfour

Commented by ajfour last updated on 24/Aug/17

T h e a r e a f r o m π / 4 t o 3 π / 4 i s

n e g a t i v e .

Commented by Joel577 last updated on 25/Aug/17

U n d e r s t o o d . T h a n k y o u v e r y m u c h

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