Recent content by jiasyuen

Use Gauss Chebyshev formula with $n=3$ to approximate the value of the integral. $$\int \frac{x^4}{\sqrt{1-x^2}}dx$$ from -1 to 1. Also compare the result with true value, where the zeros and the corresponding weights of the following simple set of orthogonal polynomial is given as below...

Can anyone actually suggest some good books for any topics for me? I've already finished 'Core Maths for Advanced Level' By L.Bostock and S.Chandler. I should move on to a higher level. Because I realize what I learn is just the basic of Mathematics.

$$\int \frac{3x-4}{x(1-x)}dx$$ $$=\int \frac{-4}{x}dx-\int \frac{1}{1-x}dx$$ $$=-4\int \frac{1}{x}dx-\int\frac{1}{1-x}dx$$ $$=-4\ln\left | x \right |-\ln \left | 1-x \right |+c$$ $$\ln \frac{x^4}{\left | 1-x \right |}+c$$ But the correct answer is $$\ln \frac{\left | 1-x \right |}{x^4}+c$$...

For example, $$\int \frac{e^x}{3e^x-1}dx$$, Should I write my answer in this $$\frac{1}{3}\ln (3e^x-1)+c$$ or $$\frac{1}{3}\ln \left | 3e^x-1 \right |+c$$ ?

Can anyone show it ? Thanks.

How to prove it by not using L'Hopital rule ?

Teach me how to simplify. $$\frac{9}{\sqrt{3}\pi \left [ (\frac{3\sqrt{3}}{2\pi})^\frac{1}{3} \right ]^2}$$

How to prove that $$\lim_{{h}\to{0}}\frac{x^h-1}{h}=\ln\left({x}\right)$$

$$\frac{9}{\sqrt{3}\pi\left [ (\frac{90\sqrt{3}}{\pi})^\frac{1}{3} \right ]^2}$$. How to simplify it into $$3^{\frac{1}{2}}(\frac{4}{\pi})^{\frac{1}{3}}m$$ ?

I'm stupid. Stuck at here :(

I've no idea.

$$\frac{dr}{dt}=\frac{9}{\sqrt{3}\pi r^2}$$ How to proceed?

Do you mean this $$\frac{dv}{dr}=\sqrt{3}\pi r^2$$?

$$v=\frac{\sqrt{3}}{3}\pi r^3$$ How to proceed from here? How to 'differentiate both sides of the formula with respect to time t' ?

$$\tan 30=\frac{r}{h}$$ $$\frac{\sqrt{3}}{3}=\frac{r}{h}$$ $$r=\frac{h}{sqrt{3}}$$ Am I correct? How to proceed?