Tích phân phụ thuộc tham số:Parameter related integrals.

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There is a discontinuity point which should be omiited at x = 0 because

Discontinuity: điểm gián đoạn, omitte: bỏ qua, consider: xem xét.

Solution:

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Fact: kết quả, sự thật.

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Factoring integrand in rational fractions – phân tích hàm dưới dấu tích phân thành phân thức hữu tỉ.

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Regard: cho, xem như.

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Code Matlaba = input('nhap vao can duoi a: ');b = input('nhap vao can tren b: ');x = a:10^(-3):b;y = log(1 + (sin(x)).^2 );y1 = 0; for i = 2 : 1 : length(x) - 1 y1 = y1+y(i);end I = ((b - a)/length(x))*( y1 + y(1)/2 + y(length(x))/2)I = pi*log(2^(1/2) + 1) - pi*log(2) = 0.5913

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(convergent: hội tụ, uniformly convergent: hội tụ đều, consider: xem xét.

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Even function: hàm số chẵn.

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Integral dependent parameter: tích phân phụ thuộc tham số.

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Therefore we can derivative integrand with respect to parameter y. Do đó ta có thể lấy đạo hàm theo tham số y dưới dấu tích phân:

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Therefore we can derivative integrand with respect to parameter y.

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Solution of homogeneous differential equation is form:

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Solution of homogeneous differential equation is form: nghiệm của pt vi phân thuần nhất có dạng.

Particular solution: nghiệm riêng.

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Solution: For , we have the following Fourier series (chuỗi Fourier):

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Now let's apply Parseval's identity (đẳng thức Parseval):

Performing now the change of variables we have

Solution:

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Solution:

This gives us:

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This can be divided into two individual (riêng biệt) sums that need to be calculated:

Since , we have C = 0, so the solution to this sum is

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Solution:

Solution:

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Solution:

Solution:

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Solution:

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Another elementary approach – 1 phương pháp tiếp cận sơ cấp khác:

Wrong:

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Solution:

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Solution:

Solution:

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Solution:

Theo quy nạp ta được:

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When two functions f and g are decomposed into power series around the same center c, the power series of the sum or difference of the functions can be obtained by termwise addition and subtraction. That is, if:

Then it is easy to show that I(0) = 0, and what we want to calculate is now the value of I(1).

Differentiating I(α), we have

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Thus integrating both side with respect to α from 0 to 1 and partially applying the substitution , we have

Here, the last term of RHS is again integral of a rational function. So we can evaluate it. Or you

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can use the taylor expansion

Thus combining (1), (3) and (4), we have

Therefore it remains to evaluate . But using (2) again, we have

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completing the proof.

Using this formula, we can obtain some special values for this integral. For example,

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