Integrals Flashcards
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| 458705322 | ∫c f(x) dx | c ∫f(x) dx | |
| 458705323 | ∫k dx | kx + C | |
| 458705324 | ∫xⁿ dx | xⁿ⁺¹ ÷ [n + 1] + C where n ≠ -1 | |
| 458705325 | ∫[1 ÷ x] dx | ln|x| +C | |
| 458705326 | ∫eˣ dx | eˣ + C | |
| 458705327 | ∫aˣ dx | [aˣ ÷ ln(a)] + C | |
| 458705328 | ∫sin(x) dx | -cos(x) + C | |
| 458705329 | ∫cos(x) dx | sin(x) + C | |
| 458705330 | ∫sec²(x) dx | tan(x) + C | |
| 458705331 | ∫csc²(x) dx | -cot(x) + C | |
| 458705332 | ∫[sec(x) tan(x)] dx | sec(x) + C | |
| 458705333 | ∫[csc(x) cot(x)] dx | -csc(x) + C | |
| 458705334 | ∫tan(x) dx | ln|sec(x)| + C ∫[sin(x) sec(x)] dx, let u = cos(x) ∴ dx = [-1 ÷ sin(x)] × du sin(x) factor cancels and it becomes ∫[1 ÷ u] du | |
| 458705335 | ∫sec(x) dx | ln|[sec(x) + tan(x)]| + C ∫sec(x)[sec(x) + tan(x) ÷ sec(x) + tan(x)] dx = ∫[sec²(x) + sec(x)tan(x)] ÷ [sec(x) + tan(x)] dx let u = sec(x) + tan(x) ∴ dx = [1 ÷ [sec(x) tan(x) + sec²(x)] du the equation becomes ∫[1 ÷ u] du | |
| 458705336 | ∫[1 ÷ (x² + 1)] dx | arctan(x) + C | |
| 458705337 | ∫[1 ÷ √(1 - x²)] dx | arcsin(x) + C | |
| 458705338 | Riemann Sum | lim (n→∞) ∑ f(xᵢ) ∆x where ∆x = (b-a) ÷ n xᵢ = a + i ∆x | |
| 458705339 | ∑ i | [n(n + 1)] ÷ 2 | |
| 458705340 | ∑ i² | [n(n + 1)(2n + 1)] ÷ 6 | |
| 458705341 | ∑ i³ | ([n(n + 1)] ÷ 2)² | |
| 458705342 | fₐᵥ | [1 ÷ (b - a)] ∫f(x) dx from a to b | |
| 458705343 | ∫(a to b) f(x)g'(x)dx | [f(x)g(x) (a to b)] - [∫(a to b) f'(x)g(x)dx | |
| 458705344 | ∫sinⁿ(x)cosᵐ(x)dx where m is odd | extract one factor of cos(x), turn remaining into factors of sin using cos²(x) = 1-sin²(x), then let u=sin(x) | |
| 458705345 | ∫sinⁿ(x)cosᵐ(x)dx where n is odd | extract one factor of sin(x), turn remaining into factors of cos using sin²(x) = 1-cos²(x), then let u=cos(x) | |
| 458705346 | ∫tanⁿ(x)secᵐ(x)dx where m is even | extract one factor of sec²(x), turn remaining into factors of tan using sec²(x) = 1-tan²(x), then let u=tan(x) | |
| 458705347 | ∫tanⁿ(x)secᵐ(x)dx where n is odd | extract one factor of sec(x)tan(x), turn remaining into factors of sec using tan²(x) = sec²(x)-1, then let u=sec(x) | |
| 458705348 | ∫√(a²-x²)]dx | x = a sinθ | |
| 458705349 | ∫√(a²+x²)]dx | x = a tanθ | |
| 458705350 | ∫√(x²-a²)]dx | x = a secθ | |
| 458705351 | R(x) ÷ [(a₁x + b)(a₂x + b)] | [A ÷ (a₁x + b)] + [B ÷ (a₂x + b)] | |
| 458705352 | R(x) ÷ [(ax + b)²] | [A ÷ (ax + b)] + [B ÷ (ax + b)²] | |
| 458705353 | R(x) ÷ [ax² + bx + c] | (Ax + B) ÷ (ax² + bx + c) | |
| 458705354 | Trapezoidal Rule for Integral Approximation | (∆x/2) [f(x₀) + 2f(x₁) + ... + 2f(xᵢ₋₁) + f(xᵢ)] | |
| 458705355 | Simpson's Rule for Integral Approximation | (∆x/3) [f(x₀) + 4f(x₁) + 2f(x₂) + 4f(x₃) + ... + 4f(xᵢ₋₁) + f(xᵢ)] | |
| 458705356 | convergence of ∫[1 ÷ xᵖ]dx | convergent for p > 1 divergent for p ≤ 1 | |
| 458705357 | Mᵧ | ρ ∫(a to b) x f(x) dx | |
| 458705358 | Mᵪ | ρ ∫(a to b) ½ [f(x)]² dx | |
| 458705359 | x̄ | (1/A) ∫(a to b) x [f(x) - g(x)] dx | |
| 458705360 | ȳ | (1/A) ∫(a to b) ½ {[f(x)]² - [g(x)]²} dx |