Derivative and integral

[steve.b]

I really need help because I am going to do my math exam tomorrow and I can't solve the tasks from the mock exam. Any help would be really appreciated!

1. Where is the following function convex/concave? +infection points
\(\displaystyle y=\frac{1}{\ln(x)}\)

2. Solve the following equation on the set of complex numbers?
\(\displaystyle (2\i+1)z^4-(16\i+8)(1+\sqrt{3}\i)=0\)

3.What is the integral of the following function?
\(\displaystyle f(x)=\ln(x+\sqrt{1+x^2})\)

4.How to determine the following acr length?
\(\displaystyle y=\ln(x)\), \(\displaystyle \left[\sqrt{3},\sqrt{8}\right]\)

5.Determine the surface of revolutionif we rotate the function around the x-axis!
\(\displaystyle y=e^{-x}\), \(\displaystyle \left[0,\infty\right[\)

Please help me!
Thank you very much in advance!
Happy new year!

 

[galactus]

1. Where is the following function convex/concave? +infection points
\(\displaystyle y=\frac{1}{\ln(x)}\)

It is concave up when \(\displaystyle f''(x)>0\), and concave down when \(\displaystyle f''(x)<0\)

If \(\displaystyle f'(x)\) is increasing, then it is concave up. If \(\displaystyle f'(x)\) is decreasing, then it is concave down.

\(\displaystyle f'(x)=\frac{-1}{x(ln(x))^{2}}\)

\(\displaystyle f''(x)=\frac{ln(x)+2}{x^{2}(ln(x))^{3}}\)

To find where it changes concavity, find the inflection point. \(\displaystyle f''(x)=0\)

Set the second derivative equal to 0 and solve for x.

2. Solve the following equation on the set of complex numbers?
\(\displaystyle (2i+1)z^4-(16i+8)(1+\sqrt{3}i)=0\)

\(\displaystyle z^{4}=\frac{(16i+8)(1+\sqrt{3}i)}{2i+1}=8+8\sqrt{3}i\)

3.What is the integral of the following function?
\(\displaystyle f(x)=\ln(x+\sqrt{1+x^2})\)

Note, this is an identity equivalent to \(\displaystyle sinh^{-1}(x)\)

Thus, it is the same as asking \(\displaystyle \int sinh^{-1}(x)dx\)

This can be looked up in a table of integrals.

Have you covered integration by parts?. It may be about as easy an any method for this one.

Let \(\displaystyle u=ln(x+\sqrt{1+x^{2}}), \;\ dv=dx, \;\ du=\frac{1}{\sqrt{x^{2}+1}}dx, \;\ v=x\)

This gives:

\(\displaystyle \underbrace{xln(x+\sqrt{1+x^{2}})}_{\text{uv}}-\int\underbrace{\frac{x}{\sqrt{x^{2}+1}}}_{\text{vdu}}dx\)

\(\displaystyle xln(x+\sqrt{1+x^{2}})-\sqrt{x^{2}+1}\)

4.How to determine the following acr length?
\(\displaystyle y=\ln(x)\), \(\displaystyle \left[\sqrt{3},\sqrt{8}\right]\)

Use the arc length formula: \(\displaystyle \int_{a}^{b}\sqrt{1+[y']^{2}}dx\)

\(\displaystyle y'=\frac{1}{x}\)

So, we integrate:

\(\displaystyle \int_{\sqrt{3}}^{\sqrt{8}}\sqrt{1+\frac{1}{x^{2}}}dx\)

\(\displaystyle \int_{\sqrt{3}}^{\sqrt{8}}\frac{\sqrt{1+x^{2}}}{x}dx\)

Can you handle this integration?.

 

[steve.b]

Thank you very much!
That was very kind of you.
Unfortunatelly I can't handle the last integral.
Happy new year!

 

[galactus]

Well, let's do it then. How would you start?. A substitution of some sort would be a good idea.