Advice 1: How to find the integral

The concept of the integral is directly related to the notion of the integral of the function. In other words, to find the integral of the specified function, you need to find the function to which the source would be derivative.
Instruction
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Integral refers to the concepts of mathematical analysis and graphically represents the area of a curvilinear trapezoid, bounded on the x-axis limit integration points. Find the integral of the function is much more complicated than finding its derivative.
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There are several methods of evaluating of indefinite integrals: direct integration, the introduction under the sign of the differential, the substitution method, integration by parts, substitution Weierstrass, the theorem of Newton-Leibnitz etc.
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Direct integration involves bringing through simple transformations of the original integral to a tabular value. Example:∫dy/(sin2y·cos2y) = ∫(cos2y + sin2y)/(sin2y·cos2y)dy = ∫dy/sin2y + ∫dy/cos2y = -ctgy + tgy + C.
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Method of introducing, under the sign of the differential or the change of variable is a new variable. The original integral is reduced to a new integral which can be converted to the tabular view by a direct integration:suppose we have the integral ∫f(y)dy = F(y) + C and a variable v = g(y), then:∫f(y)dy -> ∫f(v)dv = F(v) + C.
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Keep in mind some simple substitutions to ease working with this method:dy = d(y + b) ydy = 1/2·d(y2 + b);sinydy = - d(cosy);cosydy = d(siny).
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Пример:∫dy/(1 + 4·y2) = ∫dy/(1 + (2·y) 2) = [dy -> d(2·y)] = 1/2·∫d(2·y)/(1 + (2·y) 2) = 1/2·arctg2·the + C.
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Integration by parts is performed according to the following formula:∫udv = u·v - ∫vdu.Example:∫y·sinydy = [u = y; v = siny] = y·(cosy) – ∫(-cosy)dy = -y·cosy + siny + C.
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Definite integral in most cases is the theorem of Newton-Leibnitz:∫f(y)dy on the interval [a; b] is equal to F(b) – F(a).Example: Find ∫y·sinydy on the interval [0; 2π]:∫y·sinydy = [u = y; v = siny] = y·(cosy) – ∫(-cosy)dy = (-2π·cos2π + sin2π) – (-0·cos0 + sin0) = -2π.

Advice 2: How to calculate the integral of the function

Integral calculus is the part of mathematical analysis, basic concepts which primitive function and integral, its properties and calculation methods. The geometric meaning of these calculations – finding the area of curvilinear trapezoid bounded by the limits of integration.
Instruction
1
As a rule, the evaluation of the integral boils down to, to bring the integrand to the tabular view. There are many table-valued integrals, which facilitates the solution of such problems.
2
There are several ways to bring the integral to a convenient form: direct integration, integration by parts, the substitution method, the introduction under the sign of the differential, the Weierstrass substitution, etc.
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The method of direct integration is the successive reduction of the integral to the tabular view by using elementary transformations:∫соѕ2 (x/2)DX = 1/2•∫(1 + cos x)DX = 1/2•∫DX + 1/2•∫cos xdх = 1/2•(x + sin x) + C, where C is a constant.
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Integral has many possible values based on the properties of the integral, namely the presence of summable constants. Thus, it was found in the sample solution is shared. The private solution of the integral is called the total at a certain constant value, for example, With=0.
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Integration by parts is used when the integrand is a product of algebraic and transcendental functions. The method formula:∫udv = u•v - ∫vdu.
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Because the position of the multipliers in the product do not matter, as the function u to choose one part of the expression, which after differentiation is simplified. Example:∫x·ln xdx = [u=ln x; v=x; dv=xdx] = x2/2·ln x – ∫x2/2·dx/x = x2/2·ln x – x2/4 + C.
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The introduction of the new variable is a technique for method lookup. This changes and she integrand, and its argument:∫x·√(x - 2)dx = [t=x-2 → x = t2+2 → dx=2·tdt] = ∫(t2 + 2)·t·2·tdt = ∫(2·t^4 + 4·t2)dt = 2·t^5/5 + 4·t3/3 + C = [x=t2+2] = 2/5·(x - 2)^(5/2) + 4/3·(x - 2)^(3/2) + C.
8
Method of introducing, under the sign of the differential involves the transition to new functions. Let ∫f(x) = F(x) + C and u = g(x), then ∫f(u)du = F(u) + C [g’(x) = dg(x)]. Example:∫(2·x + 3)2dx = [dx = 1/2·d(2·x + 3)] = 1/2·∫(2·x + 3)2d(2·x + 3) = 1/6·(2·x + 3)3 + C.
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