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

&lt;br /&gt; &lt;br /&gt; What&amp;#039;s the problem? -1 &amp;lt; 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 doesn't make sense at all since [0;-1] is not an interval?\[ \int_{-1}^{0} f(x) \, \mathrm{d}x \]/requires that 0&amp;lt;-1 ..&lt;br /&gt; Please help i am VERY confused.&lt;br /&gt;

&lt;br /&gt; &lt;br /&gt; Some LaTeX tips. &lt;br /&gt; You&amp;#039;re putting in way more symbols than you actually need - extra brackets and slashes.&lt;br /&gt; Use a single pair of LaTeX delimiters for an entire equation, rather than breaking it up into multiple LaTeX expressions.&lt;br /&gt; &lt;br /&gt; Instead of writing this: \[ \int_{a}^{b} f(x) \, \mathrm{d}x \], you can write it much more simply this way: $\int_a^b f(x)~dx$&lt;br /&gt; &lt;div class=&quot;bbCodeBlock bbCodeBlock--screenLimited bbCodeBlock--code&quot;&gt; &lt;div class=&quot;bbCodeBlock-title&quot;&gt; &lt;i class=&quot;fa--xf fal fa-code &quot;&gt;&lt;svg xmlns=&quot;http://www.w3.org/2000/svg&quot; role=&quot;img&quot; aria-hidden=&quot;true&quot; &gt;&lt;use href=&quot;/data/local/icons/light.svg?v=1775307811#code&quot;&gt;&lt;/use&gt;&lt;/svg&gt;&lt;/i&gt; Code: &lt;/div&gt; &lt;div class=&quot;bbCodeBlock-content&quot; dir=&quot;ltr&quot;&gt; &lt;pre class=&quot;bbCodeCode&quot; dir=&quot;ltr&quot; data-xf-init=&quot;code-block&quot; data-lang=&quot;&quot;&gt;&lt;code&gt;$\int_a^b f(x)~dx$&lt;/code&gt;&lt;/pre&gt; &lt;/div&gt; &lt;/div&gt;&lt;br /&gt; Or instead of this: \[ \int_{a}^{b} f(x) \, \mathrm{d}x \]=-\[ \int_{b}^{a} f(x) \, \mathrm{d}x \]&lt;br /&gt; You can write this:&lt;br /&gt; $\int_a^b f(x)~dx = -\int_b^a f(x)~dx$&lt;br /&gt; &lt;div class=&quot;bbCodeBlock bbCodeBlock--screenLimited bbCodeBlock--code&quot;&gt; &lt;div class=&quot;bbCodeBlock-title&quot;&gt; &lt;i class=&quot;fa--xf fal fa-code &quot;&gt;&lt;svg xmlns=&quot;http://www.w3.org/2000/svg&quot; role=&quot;img&quot; aria-hidden=&quot;true&quot; &gt;&lt;use href=&quot;/data/local/icons/light.svg?v=1775307811#code&quot;&gt;&lt;/use&gt;&lt;/svg&gt;&lt;/i&gt; Code: &lt;/div&gt; &lt;div class=&quot;bbCodeBlock-content&quot; dir=&quot;ltr&quot;&gt; &lt;pre class=&quot;bbCodeCode&quot; dir=&quot;ltr&quot; data-xf-init=&quot;code-block&quot; data-lang=&quot;&quot;&gt;&lt;code&gt;$\int_a^b f(x)~dx = -\int_b^a f(x)~dx$&lt;/code&gt;&lt;/pre&gt; &lt;/div&gt; &lt;/div&gt;