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 doesn't make sense at all since [0;-1] is not an interval?[itex]\[ \int_{-1}^{0} f(x) \, \mathrm{d}x \]/$requires that 0<-1 ..<br /> Please help i am VERY confused.<h2>Homework Equations</h2><br /> mentioned above<h2>The Attempt at a Solution</h2><br /> mentioned above$[/itex]

 

[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 doesn't make sense at all since [0;-1] is not an interval?[itex]\[ \int_{-1}^{0} f(x) \, \mathrm{d}x \]/$requires that -1<0 ..<br /> Please help i am VERY confused.<br /> <br /> <br /> <h2>Homework Equations</h2><br /> mentioned above<br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> mentioned above$[/itex]

[itex]$<br /> What's the problem? -1 < 0 is certainly true! Why would you think otherwise?$[/itex]

 

[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 doesn't make sense at all since [0;-1] is not an interval?[itex]\[ \int_{-1}^{0} f(x) \, \mathrm{d}x \]/$requires that 0<-1 ..<br /> Please help i am VERY confused.$[/itex]

[itex][itex] <br /> Some LaTeX tips. <br /> You're putting in way more symbols than you actually need - extra brackets and slashes.<br /> Use a single pair of LaTeX delimiters for an entire equation, rather than breaking it up into multiple LaTeX expressions.<br /> <br /> Instead of writing this: $\[ \int_{a}^{b} f(x) \, \mathrm{d}x \]$, you can write it much more simply this way: $\int_a^b f(x)~dx$<br /> <div class="bbCodeBlock bbCodeBlock--screenLimited bbCodeBlock--code"> <div class="bbCodeBlock-title"> <i class="fa--xf fal fa-code "><svg xmlns="http://www.w3.org/2000/svg" role="img" aria-hidden="true" ><use href="/data/local/icons/light.svg?v=1776459805#code"></use></svg></i> Code: </div> <div class="bbCodeBlock-content" dir="ltr"> <pre class="bbCodeCode" dir="ltr" data-xf-init="code-block" data-lang=""><code>$\int_a^b f(x)~dx$</code></pre> </div> </div><br /> Or instead of this: $\[ \int_{a}^{b} f(x) \, \mathrm{d}x \]$=-$\[ \int_{b}^{a} f(x) \, \mathrm{d}x \]$<br /> You can write this:<br /> $\int_a^b f(x)~dx = -\int_b^a f(x)~dx$<br /> <div class="bbCodeBlock bbCodeBlock--screenLimited bbCodeBlock--code"> <div class="bbCodeBlock-title"> <i class="fa--xf fal fa-code "><svg xmlns="http://www.w3.org/2000/svg" role="img" aria-hidden="true" ><use href="/data/local/icons/light.svg?v=1776459805#code"></use></svg></i> Code: </div> <div class="bbCodeBlock-content" dir="ltr"> <pre class="bbCodeCode" dir="ltr" data-xf-init="code-block" data-lang=""><code>$\int_a^b f(x)~dx = -\int_b^a f(x)~dx$</code></pre> </div> </div>[/itex][/itex]