2019年5月18日星期六

全新 Fergus Sir DSE 2020 / DSE 2021 數學 CORE / M2 備戰課程


數學 Core
M2 (代數&微積分)
MC 技巧特訓
傳授必學砌數技巧
A1 A2 概念重構
深入教授微積分技術
拆解 Section B killer 題
fx3650P 輕鬆屈機

培訓考試技巧,重建答題框架
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Everyday Math 每日數學 之 Quadratic Equation and Graph 二次方程及圖像 #20190515 (+Solution)


Answer : D

2019年5月17日星期五

Everyday Math 每日數學 之 Integration #20190517 (+Solution)





(a)
$ e^x=\frac{1}{t-1} $
$ x=-\ln(t-1) $
$ dx=-\frac{1}{t-1}dt $
$ \int \frac{1}{e^{4x}+e^{3x}}dx $
$ =\int \frac{1}{\left(\frac{1}{t-1}\right)^{4}+\left(\frac{1}{t-1}\right)^{3}}\cdot -\frac{1}{t-1}dt $
$ =-\int \frac{1}{\left(\frac{1}{t-1}\right)^{3}+\left(\frac{1}{t-1}\right)^{2}} dt $
$ =-\int \frac{(t-1)^3}{1+t-1} dt $
$ =-\int \frac{t^3-3t^2+3t-1}{t} dt $
$ =-\frac{1}{3}t^3+\frac32t^2-3t+\ln|t|+C $
$ =-\frac{1}{3}\left( \frac{1}{e^x}+1\right)^3+\frac32\left( \frac{1}{e^x}+1\right)^2-3\left( \frac{1}{e^x}+1\right)+\ln\left| \frac{1}{e^x}+1\right|+C $

(b)

Let $ u=e^x $ . Then $ du=e^x dx $ .
$ \int \frac{1}{u^5+u^4}du $
$ =\int \frac{1}{e^{5x}+e^{4x}}\cdot e^xdx $
$ =\int \frac{1}{e^{4x}+e^{3x}}dx $
$ =-\frac{1}{3}\left( \frac{1}{e^x}+1\right)^3+\frac32\left( \frac{1}{e^x}+1\right)^2-3\left( \frac{1}{e^x}+1\right)+\ln\left| \frac{1}{e^x}+1\right|+C $ ( by (a) )
$ =-\frac{1}{3}\left( \frac{1}{u}+1\right)^3+\frac32\left( \frac{1}{u}+1\right)^2-3\left( \frac{1}{u}+1\right)+\ln\left|\frac{1}{u}+1\right|+C $

(c)

$ \int _1^e \frac{5u^5+4u^4}{(u^5+u^4)^2}du $
$ =\int _1^e \frac{(5u^4+4u^3)u}{(u^5+u^4)^2}du $
$ =\int _{u=1}^{u=e} \frac{u}{(u^5+u^4)^2}d(u^5+u^4) $
$ =\int _{u=1}^{u=e} u(u^5+u^4)^{-2}d(u^5+u^4) $
$ =-\int _{u=1}^{u=e} ud(u^5+u^4)^{-1} $
$ =- \left[ u(u^5+u^4)^{-1}\right]_{u=1}^{u=e} +\int _{u=1}^{u=e} (u^5+u^4)^{-1}du $
$ =- \left[ (u^4+u^3)^{-1}\right]_{u=1}^{u=e} +\left[-\frac{1}{3}\left( \frac{1}{u}+1\right)^3+\frac32\left( \frac{1}{u}+1\right)^2-3\left( \frac{1}{u}+1\right)+\ln\left| \frac{1}{u}+1\right|\right]_{u=1}^{u=e} $
$ =- (e^4+e^3)^{-1} -\frac{1}{3}\left( \frac{1}{e}+1\right)^3+\frac32\left( \frac{1}{e}+1\right)^2-3\left( \frac{1}{e}+1\right)+\ln\left( \frac{1}{e}+1\right)+\frac{19}{6}-\ln 2 $

2019年5月14日星期二

M2 Review - Chapter 13 Vector












M2 Review - Chapter 12 System of Linear Equation





M2 Review - Chapter 11 Matrix









M2 Review - Chapter 10 Application of Integration




M2 Review - Chapter 9 Integration











M2 Review - Chapter 8 Curve Sketching






M2 Review - Chapter 7 Application of Differentiation




M2 review - Chapter 6 Differentiation




M2 Review - Chapter 5 d/dx byFirst Principle