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I've may been asking the really stupid question.

The below general formula.

$$ \int_{a }^{b } f \left( x \right) \,dx =-\int_{b }^{a } f \left( x \right) \,dx $$

For instance,

$$ \int_{1 }^{3 } x \,dx = \left[ \frac{ 1 }{ 2 } x ^{2} \right]_{1}^{3} $$

$$ = \frac{1}{2} \left( 3 ^{2} -1 ^{2} \right) $$

$$ = \frac{ 1 }{ 2 } \left( 9-1 \right) =4 $$

And as we reverse the limits of the integration,

$$ \int_{3 }^{1 } x \,dx =\left[ \frac{ 1 }{ 2 } x ^{2} \right]_{3}^{1} $$

$$ = \frac{1}{2} \left( 1 ^{2} -3 ^{2} \right) $$

$$ = \frac{1}{2} \left( 1-9 \right) $$

$$ = \frac{ -8 }{ 2 } =-4 ~~ \leftarrow~~ \text{Only the sign is opposite} $$

I know that a return value of an integration is given by summing up an each infinitesimal area of a infinitesimal rectangle. In this integration for example, the width of the rectangle is $~ dx ~$ and the height of it is given by $~ x ~$

So even the limits is reversed , the result should be same but of course from the formula, that claim is disvalidated.

Why?

electrical apprentice
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    In effect the "width" of the rectangles, given by subtracting the "earlier" $x$ value from the "later" one, is negative when the limits are in the wrong order. One reason why this convention is convenient (among others) is that it means $\int_a^c f(x) dx + \int_c^b f(x) dx = \int_a^b f(x) dx$, regardless of how $a,b,c$ are ordered. This wouldn't be true if we defined $\int_a^b f(x) dx$ to be shorthand for $\int_{\min { a,b }}^{\max { a,b }} f(x) dx$. – Ian Jul 25 '21 at 01:52
  • Integration gives us an oriented or signed area. Reversing the limits of integration reverses the orientation and the sign. – user317176 Jul 25 '21 at 01:59

2 Answers2

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According to the fundamental theorem of calculus (or Newton-Leibniz rule): $$\int_a^bf(x)\, dx=F(x)|_a^b=F(b)-F(a)=-[F(a)-F(b)]=-F(x)|_b^a=-\int_b^af(x)\,dx.$$

Looking at the geometric meaning, we see that the the area function is simply an antiderivative of the original function. It is stated that for small $h$, the two values are equal: $$A(x+h)-A(x)=f(x)\cdot h.$$ So, it is a matter of which value $A(x+h)$ or $A(x)$ is greater, which depends on the function $f(x)$.

farruhota
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There are a few ways of looking at this. First, remember that (Riemann) integrals are signed areas, so that will play a role in our reasoning.

Reason $1$. It makes the formulas work out. One formula that we have for integrals is $$ \int_a^b f(x)dx+\int_b^c f(x)dx=\int_a^cf(x)dx. $$ If you try this when $c=a$, you get $$ \int_a^bf(x)dx+\int_b^af(x)dx=\int_a^af(x)dx=\int_b^bf(x)dx=0. $$ By solving for one integral in terms of the other, we conclude that $$ \int_b^af(x)dx=-\int_a^bf(x)dx. $$

Reason $2$. (Following Ian's comment). Since the rectangles that we use are signed areas, we use the convention that the positive directions are to the right and upwards. In other words, the area of a rectangle is $$ (\text{distance to the right})\times(\text{distance upwards}). $$ When $f$ is negative, the distance upwards is negative, giving a negative integral. When considering the integral $\int_b^a f(x)dx$ with $b\geq a$, the rectangles are formed from right to left, so their bases are from right to left, i.e., the "distance to the right" is negative. This reverses the sign of the integral.

Michael Burr
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