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Let the root be α. Since the coefficients are real, roots are in conjugate pairs. Hence the other root is α* (conjugate). Using the product of roots \begin{align*}\alpha \times \overline{\alpha} &= \frac{1}{1}\\|\alpha|^2 &=1\\|\alpha| &=1 && \text{since } |\alpha| >...

okay I'll assume you mean prove 2n < 3n. First note that you don't need induction to prove this – you can simply raise both sides of 2 < 3 to the power of n (as both sides are greater than zero & xn is strictly increasing for x > 0). However, using induction: 1. The base case is obvious...

well you can let f(x) = 5/2 which satisfies all of the properties then \begin{align*}\int_{0}^{50} f(x) \, \mathrm{d}x &= 50 \times \frac{5}{2}\\&=125\end{align*}

I used texshop (latex) with the pstricks package (completely free) latex pstricks

I think the answer is: \frac{1}{2}\left [x \right ] \left ( \left [ x \right ] - 2 \right )

Re: MX2 Integration Marathon \begin{align*}I &= \int^{\sqrt{\ln(3)}}_{\sqrt{\ln(2)}} \frac{x\sin(x^2)}{\sin(x^2) + \sin(\ln(6) - x^2)} \, \mathrm{d}x\\&= \frac{1}{2}\int^{\ln(3)}_{\ln(2)} \frac{\sin(x)}{\sin(x) + \sin(\ln(6) - x)} \, \mathrm{d}x\\&= \frac{1}{2}\int^{\ln(3)}_{\ln(2)}...

Re: MX2 Integration Marathon :cool: good job i did it a similar way: \begin{align*}\text{Set } G(t) = \int^{1}_{0} \frac{x^t-1}{\ln(x)}\, \mathrm{d}x \textup{, so that } G'(t) &= \frac{\mathrm{d} }{\mathrm{d} t}\int^{1}_{0} \frac{x^t-1}{\ln(x)} \, \mathrm{d}x\\&= \int^{1}_{0} \frac{\partial...

Re: MX2 Integration Marathon new question: find \int^{1}_{0} \frac{x^{n-1}-1}{\ln(x)} \, \mathrm{d}x, \quad n>0 or something easier \int \frac{2}{\left (\cos(x) - \sin(x) \right )^2} \, \mathrm{d}x

Wow. This is incredibly generous of you spiral. Thank you!

Re: MX2 Integration Marathon 5th line should be (note limits of last two integrals) \\\int^{\frac{\pi}{2}}_0 x\cot(x) \, \mathrm{d}x + \frac{\pi}{2}\int^{\frac{\pi}{2}}_0 \csc(x) - \cot(x) \, \mathrm{d}x + \int^{\frac{\pi}{2}}_0 (x-\pi)\cot(x) \, \mathrm{d}x\\+...

remember that mv2/r is the resultant force (horizontally); the other forces must add to give this. the same principle applies when you resolve this resultant force into its components. to convince yourself, resolve horizontally and vertically, then solve simultaneously (times vertical by...

Re: MX2 Integration Marathon yup

okay here's 5(a) Resolving forces along the bank can be tricky. If all else fails, resolve horizontally and vertically, then solve simultaneously.

Re: MX2 Integration Marathon sorry, nope (although interested to know your method) s/he's using properties of definite integrals, namely \int^a_0 f(x) \, \mathrm{d}x = \int^a_0 f(a-x) \, \mathrm{d}x also tan(pi/2 - x) = cot(x) :p

Re: MX2 Integration Marathon lol you forgot to integrate! \int^{\frac{\pi}{4}}_{0} \ln(2) \, \mathrm{d}x = \frac{\pi}{4}\ln(2) which gives the correct answer here is a really difficult one: \int^{\frac{\pi}{2}}_{0} \frac{x}{\tan(x)} \, \mathrm{d}x good luck

Re: MX2 Integration Marathon I = \int^1_0 \frac{\ln(x+1)}{1+x^2} \, \mathrm{d}x\\\\\begin{align*}\textup{Let } x = \frac{1-u}{1+u} = \frac{2}{1+u} - 1 \textup{, so } \mathrm{d}x = -\frac{2}{(1+u)^2} \, \mathrm{d}u\end{align*}\\\begin{align*}I &= \int^0_1 \frac{\ln\left ( \frac{2}{1+u}...

Re: MX2 Integration Marathon \begin{align*}\int \sin^{12}\left ( 7x \right )\cos^{3}\left ( 7x \right ) \, \mathrm{d}x &= \int \sin^{12}\left ( 7x \right )\cos\left ( 7x \right ) -\sin^{14}\left ( 7x \right )\cos\left ( 7x \right ) \, \mathrm{d}x\\&=\frac{1}{7\times13}\sin^{13}(7x) -...

Re: MX2 Integration Marathon good job :cool:. after simplification it becomes: \frac{1}{\sqrt{2}} \ln \left (1 + \sqrt{2} \right ) next: \int_0^{\frac{\pi}{2}} \frac{\cos^{2012}(x)}{\sin^{2012}(x) + \cos^{2012}(x)} \, \mathrm{d}x edit: \\\textup{Roots of } u^4 + 1...

Re: MX2 Integration Marathon \int ^{\frac{\pi}{2}}_0 \frac{\sin(x)}{1 + \sin^2(x)} \, \mathrm{d}x