# Six assorted calculus problems

#### ozgunozgur

##### New member
Last edited by a moderator:

#### MarkFL

Staff member
Hello, and welcome to MHB! In the future please make a thread for each problem so the resulting discussion doesn't become convoluted.

1.) I would begin by graphing $$\bf R$$:

I would use the washer method here for the volume of an arbitrary element:

$$\displaystyle dV=\pi(R^2-r^2)\,dx$$

Can you identify the outer radius $$R$$ and the inner radius $$r$$ of the arbitrary washer above?

• anemone

#### Fantini

MHB Math Helper
Mark, this does look like a test. I'd wait for the 48h before replying.

#### ozgunozgur

##### New member
Thanks, now I have a solution. So iis that true? Last edited by a moderator:

#### MarkFL

Staff member
Continuing where I left off, we have:

$$\displaystyle dV=\pi((1+\sin(x))^2-(1)^2)\,dx=\pi(\sin^2(x)+2\sin(x))\dx$$

Using a double-angle identiy for cosine: we may state:

$$\displaystyle \cos(2x)=1-2\sin^2(x)\implies \sin^2(x)=\frac{1-\cos(2x)}{2}$$

And so we may write:

$$\displaystyle dV=\frac{\pi}{2}(4\sin(x)-\cos(2x)+1)\,dx$$

Hence, adding up all the washers, we find:

$$\displaystyle V=\frac{\pi}{2}\int_0^{\pi} 4\sin(x)-\cos(2x)+1\,dx$$

Let's let:

$$\displaystyle u=x-\frac{\pi}{2}\implies du=dx$$

$$\displaystyle V=\frac{\pi}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 4\cos(u)+\cos(2x)+1\,dx$$

Using the even-function rule, we have:

$$\displaystyle V=\pi\int_{0}^{\frac{\pi}{2}} 4\cos(u)+\cos(2x)+1\,dx=\pi\left[4\sin(u)+\frac{1}{2}\sin(2u)+u\right]_{0}^{\frac{\pi}{2}}=\pi\left(4+\frac{\pi}{2}\right)=\frac{\pi}{2}(8+\pi)$$

It looks like you were on the right track, but you neglected to distribute $$\pi$$ to your additional integral.

For the second problem, it looks like you are using integration by parts, but your work is hard to read. I'll let someone else jump in to help with that. • anemone and topsquark

#### ozgunozgur

##### New member
Thanks, This is question 5, can you write me?

#### ozgunozgur

##### New member
Is this solution true?

#### ozgunozgur

##### New member
Regarding question 3, ln(1) = 0, so you need to find −∫10ln(x)dx−∫01ln⁡(x)dx.
An indefinite integral is xln(x)−xxln⁡(x)−x. Can you finish the question now?

According to W|A, the integral in question 4 doesn't converge. Did you type it correctly?

#### ozgunozgur

##### New member
In questipn 3, integral x(lnx-1) 1(0-1) - 0(ln0-1) = -1
In question 4, the integral does not converge.
One side is endless when it is solved as a normal integral.

#### MarkFL

Staff member
3. I would observe that the area in question may be found from:

$$\displaystyle A=\int_{-\infty}^0 e^x\,dx$$

This is an improper integral, and so I would write:

$$\displaystyle A=\lim_{t\to-\infty}\left(\int_t^0 e^x\,dx\right)=\lim_{t\to-\infty}\left(e^0-e^t\right)=1\quad\checkmark$$

• anemone

#### ozgunozgur

##### New member
3. I would observe that the area in question may be found from:

$$\displaystyle A=\int_{-\infty}^0 e^x\,dx$$

This is an improper integral, and so I would write:

$$\displaystyle A=\lim_{t\to-\infty}\left(\int_t^0 e^x\,dx\right)=\lim_{t\to-\infty}\left(e^0-e^t\right)=1\quad\checkmark$$
Can you help me for question 4? This is my last question, is that true?

#### MarkFL

Staff member
According to W|A the definite integral given in #4 does not converge.

$$\displaystyle I=\int_2^{\infty}\frac{x\sqrt{x}}{x^2-1}\,dx$$

$$\displaystyle u=\sqrt{x}\implies du=\frac{1}{2\sqrt{x}}\,dx\implies dx=2u\,du$$

We then have:

$$\displaystyle I=2\int_{\sqrt{2}}^{\infty}\frac{u^4}{u^4-1}\,du=2\int_{\sqrt{2}}^{\infty}1+\frac{1}{u^4-1}\,du$$

And then using partial fractions, we may write:

$$\displaystyle I=\frac{1}{2}\int_{\sqrt{2}}^{\infty} 4+\frac{1}{u-1}-\frac{1}{u+1}-\frac{2}{u^2+1}\,du$$

Do you agree so far?

• anemone

#### ozgunozgur

##### New member
According to W|A the definite integral given in #4 does not converge.

$$\displaystyle I=\int_2^{\infty}\frac{x\sqrt{x}}{x^2-1}\,dx$$

$$\displaystyle u=\sqrt{x}\implies du=\frac{1}{2\sqrt{x}}\,dx\implies dx=2u\,du$$

We then have:

$$\displaystyle I=2\int_{\sqrt{2}}^{\infty}\frac{u^4}{u^4-1}\,du=2\int_{\sqrt{2}}^{\infty}1+\frac{1}{u^4-1}\,du$$

And then using partial fractions, we may write:

$$\displaystyle I=\frac{1}{2}\int_{\sqrt{2}}^{\infty} 4+\frac{1}{u-1}-\frac{1}{u+1}-\frac{2}{u^2+1}\,du$$

Do you agree so far?
Yes, please write all of steps, sir.

#### MarkFL

Staff member
I'm having trouble following your written work. Can you neatly show how you would proceed?

• anemone

#### ozgunozgur

##### New member
I'm having trouble following your written work. Can you neatly show how you would proceed?
I did not solve that question, I just put it on paper. I thought I was writing legibly. :/

#### MarkFL

Staff member
My next step would be to write the improper integral as:

$$\displaystyle I=\frac{1}{2}\lim_{t\to\infty}\left(\int_{\sqrt{2}}^{t} 4+\frac{1}{u-1}-\frac{1}{u+1}-\frac{2}{u^2+1}\,du\right)$$

Can you give the anti-derivative we would use in our application of the FTOC on the definite integral in the limit?

• anemone

#### ozgunozgur

##### New member
My next step would be to write the improper integral as:

$$\displaystyle I=\frac{1}{2}\lim_{t\to\infty}\left(\int_{\sqrt{2}}^{t} 4+\frac{1}{u-1}-\frac{1}{u+1}-\frac{2}{u^2+1}\,du\right)$$

Can you give the anti-derivative we would use in our application of the FTOC on the definite integral in the limit?
Is 4u + ln|u+1| - ln|u-1| + 2arctan(u) result?
if t --&gt is infinity;
= 4t - 4kok(2) + ln|(t+1)/(kok(2)+1)| - ln|(t-1)/(kok(2)+1)| + 2arctan(t) - 2arctan(kok(2))

is t a divergent integral since it approximates to infinity?

#### MarkFL

• 