Question

Use part I of the Fundamental Theorem of Calculus to find the derivative of h(x)=∫1/x512arctan(t)dt

Answer 1

is it undefines integral or does it have an interval defined?

Answer 2

integral from 5 to 1/x, and the function is 12arctan(t)dt

Answer 3

Oh my, it's been long since I didin't deduce any calculus theorems.
But they say that practise makes the master, so:
The first part of this theorem says:
"Given the integrable function in the interval [a,b], we define F being the intergral with such extremes in the differential t. If it's continious in c belonging to the interval (a,b), then F being differentiable in c and F'(c)=F(c)".
Okay, so the leme says that:
\[be (f)inte[a,b],m \le f(x) \le M,\forall x \in \left[ a,b \right]\]
Then:
\[m(b-a) \le \int\limits_{a}^{b} f(t)dt \le M(b-a)\]
That's just what the leme says, so let's go on with the real proof:
so by definition:
\[F'(c)=\lim_{h \rightarrow 0}\frac{ F(c+h)-f(c) }{ h }\]
If h>0 then:
\[F(c+h)-F(c)=\int\limits_{c}^{c+h}f(t)dt\]
So, if I define L sub h and E sub h such that:
\[L _{h}=\inf \left\{ f(x),c \le x \le c+h \right\}\]
\[E _{h}=\sup \left\{ f(x),c \le x \le c+h\right\}\]
Now, that's applyable to the leme, so let's apply it:
\[L _{h}.h \le \int\limits_{c}^{c+h}f(t)dt \le E _{h}.h\]
then:
\[L _{h} \le \frac{ F(c+h)-F(c) }{ h } \le E _{h}\]
So if h<0, let's define:
\[L*_{h}=\inf \left\{ f(x), c+h \le x \le c \right\}\]
\[E*_{h}=\sup \left\{f(x),x+h \le x \le c \right\}\]
applying same thing:
\[L*_{h}.(-h) \le \int\limits_{c+h}^{c}f(t)dt \le E*_{h}(-h) \]
Remember from your integral calculus classes that the integral becomes negative if we flip it's extremes.
Then:
\[E*_{h}(h)\le F(c+h)-F(c) \le L*_{h}(h)\]
Then, since I said h<0, then:
\[L*_{h}\le \frac{ F(c+h)-F(c) }{ h } \le E*_{h}\]
since f is continous, then:
\[\lim_{x \rightarrow h}L _{h}=\lim_{h \rightarrow 0}E _{h}=\lim_{h \rightarrow 0}L*_{h}=\lim_{h \rightarrow 0}E*_{h}=f(c)\]
So that only means one thing:
\[F'(c)=\lim_{h \rightarrow 0}\frac{ F(c+h)-F(c) }{ h }=f(c) \]
so:
\[F'(c)=f(c)\]
And that proves the theorem.

Answer 4

Now the theorem states one thing, to find the derivative of a finite integral, we just have to plug the top number of it in the function to find the derivative of that function.
Very easy to say haha.
So let's move on the problem:
\[h(x)=\int\limits_{5}^{\frac{ 1 }{ x }}12\arctan(t)dt\]
Instead of t I'll plug 1/x:
\[h'(x)=12\arctan(\frac{ 1 }{ x })\]
And as the theorem said, that's the derivative.

Answer 5

Oh god thank you so much!!!!

Answer 6

No problem :)