# Given a function F(x)=int_0^x (t+8)/(t^3-9)dt, is F′(x) different when x<0, when x=0 and when x>0?

Derivative of an integral from 0 to x when x is negative?
Given a function $F\left(x\right)={\int }_{0}^{x}\frac{t+8}{{t}^{3}-9}dt,$, is F′(x) different when $x<0$, when $x=0$ and when $x>0$?
When $x<0$, is ${F}^{\prime }\left(x\right)=-\frac{x+8}{{x}^{3}-9}$.
... since you can't evaluate an integral going from a smaller number to a bigger number? That's what I initially thought, but when I graphed the antiderivative the intervals of increase and decrease were different from the ones in my calculation.
EDIT: Yes, I was talking about that identity. So is ${F}^{\prime }\left(x\right)=\frac{x+8}{{x}^{3}-9}$ or ${F}^{\prime }\left(x\right)=-\frac{x+8}{{x}^{3}-9}$.
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kamphundg4
Step 1
We have ${F}^{\prime }\left(x\right)=\frac{x+8}{{x}^{3}-9}$ for all $x<\sqrt[3]{9}$. The limitation is due to the fact that the integral is meaningful only when the interval doesn't contain $\sqrt[3]{9}$ and so we must consider only the interval $\left(-\mathrm{\infty },\sqrt[3]{9}\right)$ that contains 0.
If $b, one sets, by definition,
${\int }_{a}^{b}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt=-{\int }_{b}^{a}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt$
so the equality ${\int }_{a}^{c}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt={\int }_{a}^{b}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt+{\int }_{b}^{c}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt$ without restrictions on the limits of integration, provided we don't jump over points where f is not defined so that all the integrals make sense.
Step 2
This relation is what the fundamental theorem of calculus relies on. Remember that, for continuous f, there exists $\xi$ such that $\frac{1}{b-a}{\int }_{a}^{b}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt=f\left(\xi \right)$ where $\xi \in \left[a,b\right]$ if $a. Therefore, for $h>0$,
${\int }_{0}^{x+h}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt-{\int }_{0}^{x}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt={\int }_{x}^{0}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt+{\int }_{0}^{x+h}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt-{\int }_{0}^{x}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt={\int }_{x}^{x+h}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt=hf\left(\xi \right)$ for some $\xi \in \left[x,x+h\right]$, so that $\underset{h\to 0+}{lim}\frac{F\left(x+h\right)-F\left(x\right)}{h}=f\left(x\right).$.
Similarly for the limit from the left.
###### Did you like this example?
Deromediqm
Step 1
Given any interval $I\subset \mathbb{R}$ and a continuous function $f:\phantom{\rule{mediummathspace}{0ex}}I\to \mathbb{R}$ the indefinite integral of f over I is the set of all functions $F:\phantom{\rule{mediummathspace}{0ex}}I\to \mathbb{R}$ (called primitives of f) that satisfy
This set depends only on (I and) f; therefore it is allowed to denote it by .
Step 2
Given any two points a, $b\in I$, the difference $F\left(b\right)-F\left(a\right)$, computed for an arbitrary primitive F of the given f, is called the definite integral of f from a to b. This difference does not depend on the chosen primitive F, nor on the name of the variable used, but only on f, a, and b. Therefore it is allowed to denote it by , or using some other "dummy variable" instead of t. From the definition it is clear that .
Step 3
What does all of this have to do with the FTC? Well, the FTC says that when $a then
.
Here the LHS is a limit of Riemann sums (an "area"), and the RHS is a difference of end-values.