حساب انتگرال
Integrals
انتگرال معین، تغییر متغیر و ترفندهای تقارن.
ریاضی عمومی کنکور ارشد علوم کامپیوتر
کارتها · 16
If f is increasing on [a,b], then g(x)=∫_a^x f is convex, continuous, and Lipschitz, but increasing only when f≥0.
Integrate by parts to express an integral with exponent n in terms of the same integral with a smaller exponent, building a recurrence you can iterate down to a base case.
Reflecting the variable across the midpoint of the interval leaves a definite integral unchanged; adding the original to the reflected copy often collapses a hopeless integrand into something trivial.
When the unknown function sits both outside and inside an integral with a variable upper limit, differentiate both sides: the FTC turns the integral into a value of the integrand, collapsing the equation into a solvable ODE.
For continuous nonnegative f on a finite interval, the n-th-root of the integral of fⁿ converges to the maximum of f — so ∫fⁿ grows like (max f)ⁿ.
The sector swept from the origin along the unit hyperbola x=cosh t, y=sinh t has area exactly t/2 — the geometric meaning of the parameter.
Split [a,b] into n equal panels of width h=(b-a)/n and sum trapezoids: the interior nodes count twice, the two endpoints once, giving (h/2)[f0 + 2(f1+...+f_(n-1)) + fn].
The ratio (sin((2n+1)x)/2)/(sinx/2) collapses into the clean sum 1+2∑_(k=1)^ncos kx — making any integral of it trivial.
When the limits a, b of a definite integral are free, the integral is largest when [a, b] is exactly the set where the integrand is nonnegative — so set the limits at consecutive roots of f.
Build the integral from sup/inf over partitions: the lower integral pushes up from below, the upper pushes down from above. They always satisfy lower ≤ upper, and f is integrable exactly when they are equal.
Integrate a ratio of two linear sine-cosine expressions by splitting the numerator into a multiple of the denominator, a multiple of the denominator's derivative, and a constant — turning the integral into a constant, a log, and a standard half-angle integral.
A rapidly oscillating factor like sin(nx) or cos(nx) averages any fixed integrable function to zero, so ∫_a^b f(x)sin(nx) dx → 0 as n→∞ . When f' is continuous, integration by parts proves it and gives the Θ(1/n) decay rate.
Integrals of the form 1/(x·√(x²±a²)) are antidifferentiated by the reciprocal substitution x = 1/t, which collapses them to ±1/√(1±t²) — an inverse-hyperbolic (or arcsin) form. The improper version converges because the tail decays like 1/x².
Area swept by a polar curve is the sum of thin circular sectors, giving A = ½∫ r² dθ; for a region between two curves integrate ½(r_outer² − r_inner²), and for an overlap take the smaller radius on each ray.
A quadrature rule ∫_a^b f ≈ Q has degree of precision d if it is exact for every polynomial of degree ≤ d but not for degree d+1 . Taylor-expanding both sides about a point and matching terms reveals d and the leading error: the first mismatched power gives an error ∝ h^(d+2) f^((d+1))(ξ) .
For nonnegative integrands, domination decides convergence: if 0≤ f≤ g then ∫ g<∞ forces ∫ f<∞ , and ∫ f=∞ forces ∫ g=∞ .