How to determine whether an integral is convergent or divergent

[ineedhelpnow]

Determine whether the integral is convergent or divergent. Evaluate those that are convergent.

$\int_{0}^{9} \ \frac{1}{\sqrt[3]{x-1}},dx$

$\int_{-\infty}^{\infty} \ \cos\left({\pi t}\right),dt$

how do i determine whether it's conver/diver?

 

[MarkFL]

Let's look at the first one...the domain of the integrand excludes $x=1$, and so it would be wise to look at:

$$\int_a^1 (x-1)^{-\frac{1}{3}}\,dx$$

To see if this converges, we could look at the limit:

$$\lim_{t\to1}\left[\int_a^t (x-1)^{-\frac{1}{3}}\,dx\right]$$

Does this limit exist, and if so, what is its value?

 

[ineedhelpnow]

I don't really know. I haven't done limits in a loooong time. I'll have to go over them once more.

 

[Jameson]

I don't really know. I haven't done limits in a loooong time. I'll have to go over them once more.

Start by evaluating the integral with $t$ and see what you get. Then take the limit. Maybe the answer will become clearer after you integrate.

 

[ineedhelpnow]

is it $\frac{-3(a-1)^{2/3}}{2}$ ?

 

[MarkFL]

is it $\frac{-3(a-1)^{2/3}}{2}$ ?

Indeed it is! :D

So, how can you use your formula to find the value of the integral in the first problem?

 

[ineedhelpnow]

that... i do not know

 

[MarkFL]

that... i do not know

Suppose we have:

$$a<k<b$$ and $f(x)$ is integrable on $[a,b]$.

A wonderful property of definite integrals is that:

$$\int_{a}^{b} f(x)\,dx=\int_{a}^{k} f(x)\,dx+\int_{k}^{b} f(x)\,dx$$

Another identity you may need is:

$$\int_{a}^{b} f(x)\,dx=-\int_{b}^{a} f(x)\,dx$$

Can you put these together now to evaluate the first integral?

 

[ineedhelpnow]

$\int_{0}^{1} \ \frac{1}{\sqrt[3]{x-1}},dx + \int_{1}^{9} \ \frac{1}{\sqrt[3]{x-1}},dx$

is the set up correct?

 

[MarkFL]

$\int_{0}^{1} \ \frac{1}{\sqrt[3]{x-1}},dx + \int_{1}^{9} \ \frac{1}{\sqrt[3]{x-1}},dx$

is the set up correct?

Yes, good! Now apply the second identity I gave so that the second integral also has 1 as the upper limit. :D

 

[ineedhelpnow]

$$\int_{0}^{1} \ \frac{1}{\sqrt[3]{x-1}}dx - \int_{9}^{1} \ \frac{1}{\sqrt[3]{x-1}} dx$$

$$=[\frac{3 (x-1)^{2/3}}{2}]^{1}_{0} - [\frac{3 (x-1)^{2/3}}{2}]^{1}_{9}$$

$$=(0-\frac{3}{2} (-1)^{2/3}) - (0-\frac{3}{2} (8)^{2/3})$$

$$=\frac{21}{4}-\frac{3\sqrt{3}}{4}$$

i think i made a mistake with the first part of my answer. my calculator is giving me -21 instead of 21

 

[MarkFL]

$\int_{0}^{1} \ \frac{1}{\sqrt[3]{x-1}},dx - \int_{9}^{1} \ \frac{1}{\sqrt[3]{x-1}} ,dx$

$=[\frac{3 (x-1)^{2/3}}{2}]^{1}_{0} - [\frac{3 (x-1)^{2/3}}{2}]^{1}_{9}$

$=(0-\frac{3}{2} (-1)^{2/3}) - (0-\frac{3}{2} (8)^{2/3})$

$=\frac{21}{4}-\frac{3\sqrt{3}}{4}$

i think i made a mistake with the first part of my answer. my calculator is giving me -21 instead of 21

You really only need to plug into your formula:

$$I=-\frac{3}{2}\left((0-1)^{\frac{2}{3}}-(9-1)^{\frac{2}{3}}\right)=-\frac{3}{2}\left(1-4\right)=\frac{9}{2}$$

 

[ineedhelpnow]

after you found that the integral was convergent wouldn't it have been the same to just work on the original integrand that was given in the first place?

 

[MarkFL]

after you found that the integral was convergent wouldn't it have been the same to just work on the original integrand that was given in the first place?

Yes, but why integrate all over again?

 

[Prove It]

Determine whether the integral is convergent or divergent. Evaluate those that are convergent.

$\int_{0}^{9} \ \frac{1}{\sqrt[3]{x-1}},dx$

$\int_{-\infty}^{\infty} \ \cos\left({\pi t}\right),dt$

how do i determine whether it's conver/diver?

For the second one, do these limits exist? $\displaystyle \begin{align*} \lim_{x \to \infty} \cos{(x)} \end{align*}$ and $\displaystyle \begin{align*} \lim_{x \to \infty} \sin{(x)} \end{align*}$