## Callisto 4 years ago Just for practice

1. Callisto

(A) Four basic operations (i) Addition => + $$+$$ (ii) Subtraction => - $$-$$ (iii) Multiplication => \times $$\times$$ or \cdot $$\cdot$$ (iv) Division => \div $$\div$$ or / $$/$$ ** (v) plus / minus => \pm $$\pm$$ ** (vi) minus/plus => \mp $$\mp$$

2. Callisto

(B) ''Equal'' signs (i) Equal => = $$=$$ (ii) Not equal to => \ne $$\ne$$ (iii) Approximately equal to => \approx $$\approx$$ (iv) Similar to => \sim $$\sim$$ (v) Congruent to => \cong $$\cong$$

3. Callisto

(C) Inequality signs (i) less than => < $$<$$ (ii) less than or equal to => \le $$\le$$ (iii) greater than => > $$>$$ (vi) greater than or equal to => \ge $$\ge$$

4. Callisto

(D) Exponent and logarithm (i) power => x^{n} $$x^{n}$$ (ii) base => x_{n} $$x_{n}$$ (iii) square root => \sqrt{x} $$\sqrt{x}$$ (iv) nth root => \sqrt[n]{x} $$\sqrt[n]{x}$$ (v) common log => \log x $$\log x$$ (vi) common log with base n => \log_{n}x $$\log_{n}x$$ (vii) natural log => \ln x $$\ln x$$

5. Callisto

(E) Binomial expansion (i) summation => \sum_{}^{} $$\sum_{n=0}^{\infty}$$ (ii) combination => _{n}C_{r} $$_{n}C_{r}$$

6. Callisto

(F) Probability (i) Combination => _{n}C_{r} $$_{n}C_{r}$$ (ii) Permutation => _{n}P_{r} $$_{n}P_{r}$$ (iii) A∪B => A\cup B $$A\cup B$$ (iv)

7. Callisto

(F) Probability (i) Combination => _{n}C_{r} $$_{n}C_{r}$$ (ii) Permutation => _{n}P_{r} $$_{n}P_{r}$$ (iii) A∪B => A\cup B $$A\cup B$$ (iv) A∩B => A \cap B $$A \cap B$$

8. Callisto

$\int_{-2}^{2}\sqrt{4-x^2}dx$ Using trigo substitution: Let x = 2sinθ , dx = 2cosθ dθ When x = 2, θ = π/2 When x = -2, θ = -π/2 The integral becomes $\int_{-\frac{π}{2}}^{ \frac{π}{2}}\sqrt{4-(2\sin\theta)^2}(2\cos\theta d\theta)$$=4\int_{-\frac{π}{2}}^{ \frac{π}{2}}\cos^2\theta d\theta$$=4\int_{-\frac{π}{2}}^{ \frac{π}{2}}\frac{\cos(2\theta)+1}{2} d\theta$$=2\int_{-\frac{π}{2}}^{ \frac{π}{2}}(\cos(2\theta)+1)d\theta$$=2(\frac{1}{2}sin(2\theta)+\theta|_{-\frac{π}{2}}^{ \frac{π}{2}})$$=2(\frac{\pi}{2}-(-\frac{\pi}{2}))$$= 2\pi$ This is not really I want to show LOL! Suppose $$y=f(x)=\sqrt{4-x^2}$$ |dw:1383399191453:dw| It's clear that integrating the function y=f(x) from -2 to 2 is the same as finding the area of the semi-circle with radius = 2. So, immediately, we get $\int_{-2}^{2}\sqrt{4-x^2}dx=\pi(2^2) /2 = 2\pi$ Geometry has its role here :D