Thread: when is this true for integral

1. when is this true for integral

what is the condition for this statement to be true?

2. Originally Posted by szpengchao

what is the condition for this statement to be true?
They may be more general hypothesis for the Fundemental Theorem of Calculus, but I know that this works.

If g is differentiable on [a,b] and g' integrable on [a,b], then

$\displaystyle \int_{a}^{x}g'(t)dt=g(x)-g(a)$

for each $\displaystyle x \in [a,b]$

ok, cheers.

4. prove

how to prove sqrt(x) is differentiable?

5. Originally Posted by szpengchao
how to prove sqrt(x) is differentiable?
Use the definition of the derivative.

$\displaystyle f'(x)=\lim_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}$

Multilpy the numerator and denominator by the conjugate to get

$\displaystyle f'(x)=\lim_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\left( \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\ri ght)=\lim_{h \to 0}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}=\frac{1}{2\sqrt{x}}$

Good luck

6. Originally Posted by TheEmptySet
Use the definition of the derivative.

$\displaystyle f'(x)=\lim_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}$

Multilpy the numerator and denominator by the conjugate to get

$\displaystyle f'(x)=\lim_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\left( \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\ri ght)=\lim_{h \to 0}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}=\frac{1}{2\sqrt{x}}$

Good luck
I think that this works as long as $\displaystyle x \ne 0$.

Here is why: if It is differentiable at zero then this limit must exist

$\displaystyle f'(0)=\lim_{h \to 0^+}\frac{\sqrt{0+h}-\sqrt{0}}{h}=\lim_{h \to 0^+}\frac{\sqrt{h}}{h}=\lim_{h \to 0^+}\frac{1}{\sqrt{h}}=\infty$

So $\displaystyle f(x)=\sqrt{x}$ is only differentable on $\displaystyle (0,\infty)$

I think you want to show that

$\displaystyle \int_{0}^{1}\sqrt{x}dx$ is Riemann integrabole on [0,1]

Here is one method let $\displaystyle \epsilon > 0$

$\displaystyle f(x)=\sqrt{x}$ is continuous on $\displaystyle [0,1]$ so it is uniformly continuous on $\displaystyle [0,1]$

Since f is unifomly continous on [0,1] there exists a $\displaystyle \delta$ such that $\displaystyle |f(x)-f(y)|< \epsilon$

Now let $\displaystyle P=\{x_0,x_1,...x_n\}$be any partition of [0,1] such that $\displaystyle |x_{j}-x_{j-1}|<\delta$. Then by the extreme value theorem there are points $\displaystyle x_M,x_m$ are the max and min on $\displaystyle x_j,x_{j-1}$ for each j.

Since the mesh of P is less than delta we have $\displaystyle |x_M-x_m|< \delta$ so $\displaystyle |f(x_M)-f(x_m) |< \epsilon$
So the upper and lower sums are

$\displaystyle U(f,P)=\sum_{j=1}^{n}f(x_M)(x_j-x_{j-1})$
$\displaystyle L(f,P)=\sum_{j=1}^{n}f(x_m)(x_j-x_{j-1})$

$\displaystyle U(f,P)-L(f,P)=\sum_{j=1}^{n}(f(x_M)-f(x_m)(x_j-x_{j-1})< \epsilon \sum_{j=1}^{n}(x_j-x_{j-1})=\epsilon(1-0)=\epsilon$