Why Does Spivak Define Integrals with a>b as Negative?

[Andrax]

Homework Statement

so this is my first time learning about integrals , from spivak' calculus
Actual quote : the integral $\[ \int_{a}^{b} f(x) \, \mathrm{d}x \]$was defined only for a<b we now add the definition
$\[ \int_{a}^{b} f(x) \, \mathrm{d}x \]$=-$\[ \int_{b}^{a} f(x) \, \mathrm{d}x \]$ if a>b "
isn't he contradicting himself here to write$\[ \int_{a}^{b} f(x) \, \mathrm{d}x \]$ a<b is required right?so you can't just write $\[ \int_{a}^{b} f(x) \, \mathrm{d}x \]$ when yo usay "if a >b"
i tried doing problem 7 which involves the function x^3
we have $\[ \int_{-1}^{1} x^3 \, \mathrm{d}x \]$=$\[ \int_{-1}^{0} f(x) \, \mathrm{d}x \]$ + $\[ \int_{0}^{1} f(x) \, \mathrm{d}x \]$(so far everything is normal) =applying spivak's definition -$\[ \int_{-1}^{0} f(x) \, \mathrm{d}x \]$ +$\[ \int_{0}^{1} f(x) \, \mathrm{d}x \]$ why in the answer books he says this equals 0 ? this dosen't make sense at all since [0;-1] is not an interval?[itex]\[ \int_{-1}^{0} f(x) \, \mathrm{d}x \]/[itex] requires that 0<-1 ..
Please help i am VERY confused.

Homework Equations

mentioned above

The Attempt at a Solution

mentioned above

 

[harikasri]

It is an odd function.

 

[Ray Vickson]

Homework Statement

so this is my first time learning about integrals , from spivak' calculus
Actual quote : the integral $\[ \int_{a}^{b} f(x) \, \mathrm{d}x \]$was defined only for a<b we now add the definition
$\[ \int_{a}^{b} f(x) \, \mathrm{d}x \]$=-$\[ \int_{b}^{a} f(x) \, \mathrm{d}x \]$ if a>b "
isn't he contradicting himself here to write$\[ \int_{a}^{b} f(x) \, \mathrm{d}x \]$ a<b is required right?so you can't just write $\[ \int_{a}^{b} f(x) \, \mathrm{d}x \]$ when yo usay "if a >b"
i tried doing problem 7 which involves the function x^3
we have $\[ \int_{-1}^{1} x^3 \, \mathrm{d}x \]$=$\[ \int_{-1}^{0} f(x) \, \mathrm{d}x \]$ + $\[ \int_{0}^{1} f(x) \, \mathrm{d}x \]$(so far everything is normal) =applying spivak's definition -$\[ \int_{-1}^{0} f(x) \, \mathrm{d}x \]$ +$\[ \int_{0}^{1} f(x) \, \mathrm{d}x \]$ why in the answer books he says this equals 0 ? this dosen't make sense at all since [0;-1] is not an interval?[itex]\[ \int_{-1}^{0} f(x) \, \mathrm{d}x \]/[itex] requires that -1<0 ..
Please help i am VERY confused.

Homework Equations

mentioned above

The Attempt at a Solution

mentioned above

What's the problem? -1 < 0 is certainly true! Why would you think otherwise?

 

[Andrax]

What's the problem? -1 < 0 is certainly true! Why would you think otherwise?

well i miss typed that, i mean 0<-1 laughs , well i think none is getting me here?
can someone prove that $\[ \int_{-1}^{1} x^3 \, \mathrm{d}x \]$ = 0 ? by splitting the intervals to -1 , 0 and 0 1 , that would help clear these things

 

[LCKurtz]

well i miss typed that, i mean 0<-1 laughs , well i think none is getting me here?
can someone prove that $\[ \int_{-1}^{1} x^3 \, \mathrm{d}x \]$ = 0 ? by splitting the intervals to -1 , 0 and 0 1 , that would help clear these things

If $f$ is an odd function so $f(-x) = -f(x)$ then for $a>0$,$$
\int_{-a}^af(x)\, dx =\int_{-a}^0f(x)\, dx + \int_0^a f(x)\, dx$$Let $x=-u$ in the first integral making it$$
\int_{a}^0f(-u)\, (-1)du = \int_{a}^0 -f(u)\, (-1)du = \int_a^0f(u)\, du
=-\int_0^af(u)\, du$$so this integral cancels the second integral, giving $0$.

 

[Andrax]

If $f$ is an odd function so $f(-x) = -f(x)$ then for $a>0$,$$
\int_{-a}^af(x)\, dx =\int_{-a}^0f(x)\, dx + \int_0^a f(x)\, dx$$Let $x=-u$ in the first integral making it$$
\int_{a}^0f(-u)\, (-1)du = \int_{a}^0 -f(u)\, (-1)du = \int_a^0f(u)\, du
=-\int_0^af(u)\, du$$so this integral cancels the second integral, giving $0$.

thanks everything is clear now and a little bit offtopic , wow Integral is hard compared to derivatives and limits i might switch to another book spivak became suddenly very hard to me

 

[Mark44]

Homework Statement

so this is my first time learning about integrals , from spivak' calculus
Actual quote : the integral $\[ \int_{a}^{b} f(x) \, \mathrm{d}x \]$was defined only for a<b we now add the definition
$\[ \int_{a}^{b} f(x) \, \mathrm{d}x \]$=-$\[ \int_{b}^{a} f(x) \, \mathrm{d}x \]$ if a>b "
isn't he contradicting himself here to write$\[ \int_{a}^{b} f(x) \, \mathrm{d}x \]$ a<b is required right?so you can't just write $\[ \int_{a}^{b} f(x) \, \mathrm{d}x \]$ when yo usay "if a >b"
i tried doing problem 7 which involves the function x^3
we have $\[ \int_{-1}^{1} x^3 \, \mathrm{d}x \]$=$\[ \int_{-1}^{0} f(x) \, \mathrm{d}x \]$ + $\[ \int_{0}^{1} f(x) \, \mathrm{d}x \]$(so far everything is normal) =applying spivak's definition -$\[ \int_{-1}^{0} f(x) \, \mathrm{d}x \]$ +$\[ \int_{0}^{1} f(x) \, \mathrm{d}x \]$ why in the answer books he says this equals 0 ? this dosen't make sense at all since [0;-1] is not an interval?[itex]\[ \int_{-1}^{0} f(x) \, \mathrm{d}x \]/[itex] requires that 0<-1 ..
Please help i am VERY confused.

Some LaTeX tips.
You're putting in way more symbols than you actually need - extra brackets and slashes.
Use a single pair of LaTeX delimiters for an entire equation, rather than breaking it up into multiple LaTeX expressions.

Instead of writing this: $\[ \int_{a}^{b} f(x) \, \mathrm{d}x \]$, you can write it much more simply this way: $\int_a^b f(x)~dx$

##\int_a^b f(x)~dx##

Or instead of this: $\[ \int_{a}^{b} f(x) \, \mathrm{d}x \]$=-$\[ \int_{b}^{a} f(x) \, \mathrm{d}x \]$
You can write this:
$\int_a^b f(x)~dx = -\int_b^a f(x)~dx$

## \int_a^b f(x)~dx = -\int_b^a f(x)~dx##