1

Can one prove $$\int^b_a f(t) \ dt = - \int^a_b f(t) \ dt$$ ? Similarly can one prove $$\int^a_a f(t) \ dt = 0$$ ? Is equality only by definition in both?

I've learned that $\int^b_a f(t) \ dt$ represent the equal values of the lower and upper integrals (supremum and infimum with respect to lower and upper sums of partitions). Also a partition is defined as follows: $$a = x_0 < x_1 < ... < x_n = b$$, which imply $a < b$.

So how does one make any sense of the equalities above ? Can one prove them by means of theory (please show me how), or why have people chosen these definitions ?

Shuzheng
  • 5,821

3 Answers3

3

Alternatively, if you use Riemann sums, $$\int_a^b f(x) dx = \lim_{N \to \infty} \frac{b-a}{N}\sum_{k=0}^{N-1} f\left(a+k \frac{b-a}{N}\right),$$ then swapping $a$ and $b$ reverseing the order of the sum using $k=N-j$, you will get the same expression but with a minus sign (up to a small error term that goes away in the limit).

abnry
  • 14,976
  • So what you get is $$\int_b^a f(x) dx = \lim_{N \to \infty} \frac{a-b}{N}\sum_{j=0}^{N-1} f\left(b+(N-j) \frac{a-b}{N}\right),$$ ? – Shuzheng Mar 29 '14 at 17:41
  • Distribute $(N-j)$ to get the same argument in the function as originally written. Also note the bounds should be $j=1$ to $j=N$, hence the comment about an error term that goes away with the limit. You have to add and subtract $f(a)$ and $f(b)$ respectively. – abnry Mar 29 '14 at 17:45
  • Does this "prove" the integral with these bounds ? – Shuzheng Mar 29 '14 at 18:32
  • What I am saying is that using this definition you get $\int_a^b = -\int_b^a$. – abnry Mar 29 '14 at 18:59
0

Notice that $$\int_a^b f(t)dt+\int_b^c f(t) dt=\int_a^c f(t) dt $$ thus if $c=a$ rigth hand side become zero and we are done.

$$\int_a ^a f(t)dt$$ being zero is clear from many aspect.

mesel
  • 15,125
  • But if $c = a$ then one of your integrals on left hand side is undefined ? – Shuzheng Mar 29 '14 at 15:06
  • @user111854: I think that is why we define it in that matter. But nayrb's answer is more clear. it is like why $0!=1$. – mesel Mar 29 '14 at 15:36
0

Let $F(x)=\int f(x)dx$ (Indefinite)

then your first equation is $F(b)-F(a)$ (Arbitrary constant gets cancelled).

I assume you can carry on from here. And those partitions need are true even if inequalities don't hold but we like to define it that way as integral from a to b represents area under curve from a to b. Just google geometrical significance of integral.

evil999man
  • 6,088