Here's the question you clicked on:
Callisto
Just for practice
(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 \(\mp\)
(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 \(\cong\)
(C) Inequality signs (i) less than => < \(<\) (ii) less than or equal to => \le \(\le\) (iii) greater than => > \(>\) (vi) greater than or equal to => \ge \(\ge\)
(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 \(\ln x\)
(E) Binomial expansion (i) summation => \sum_{}^{} \(\sum_{n=0}^{\infty}\) (ii) combination => _{n}C_{r} \(_{n}C_{r}\)
(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)
(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 \(A \cap B\)
\[\int_{-2}^{2}\sqrt{4-x^2}dx\] 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