Topic: Improper Integrals
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How to Calculate an Improper Integral
Improper integrals have positive or negative infinity in one or both of the bounds of the integrals. It is difficult to imagine calculating the area of a graph stretching to positive or negative infinity, but it is possible with the use of ... Read More »
Source: http://www.ehow.com/how_8470071_calculate-improper-integral.html
What is improper integral?
n. An integral having at least one nonfinite limit or an integrand that becomes infinite between the limits of integration. Read More »
Source: http://www.answers.com/topic/improper-integral
How to solve this improper integral?
The fastest and best way to do this integral is via Euler's formula e^(it) = cos(t) + i sin(t) Then Int_(0,oo) e^(-3x) sin(5x) dx = Im { Int_(0,oo) e^(-3x) e^(5ix) dx } = Im { Int_(0,oo) e^(- (3-5i)x) dx } = Im { 1/(3 - 5i) } = Im { (3+5i)/... Read More »
Source: http://answers.yahoo.com/question/index?qid=20120223154639AAlAkSM
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Improper Integrals
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as typed a) is proper ; b) has problems at x = 2 { unbounded } ; c) infinite domain and in a) the domain should be [ 1 , t ] since x is the integration variable
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Source: http://answers.yahoo.com/question/index?qid=20111016204449AAvXJJF
a and d are improper d is improper because you are allowing x to approach infinity and/or negative infinity. a is improper because the function itself goes to infinity and/or negative infinity for some value x. (i.e. when x approaches 6, th...
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Source: http://answers.yahoo.com/question/index?qid=20100906204851AA4uo2T
First let's just worry about the indefinite integral. Using partial fractions you get: (1 / (2x^2 + x)) = ( -2 / 2x + 1) ( 1 / x) (You can check this easily simply by realizing that the lcd is ((2x+1) * x and that ((-2x + 2x + 1) . (2x^2 + ...
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Source: http://answers.yahoo.com/question/index?qid=20100219174942AAafCod
Integral (1 to infinity, 1/x^3 dx ) All improper integrals must be expressed as a limit. lim Integral (1 to t, 1/x^3 dx ) t -> infinity Which is easily integrable. lim Integral (1 to t, x^(-3) dx ) t -> infinity lim [ (-1/2)x^(-2) ] {evalua...
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Source: http://answers.yahoo.com/question/index?qid=20090414120845AAL5HxT
∫ x/(x-4)³ dx = (2-x)/(x-4)² +C lim (2-x)/(x-4)² = ∞ x→4 So it diverges ..
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Source: http://answers.yahoo.com/question/index?qid=20090118135246AAjxOqn
Since I cannot integrate that well, I'll do the last two problems. 2) ∫ ln(x) dx = ∫ 1*ln(x) Letting u = ln(x) and dv = 1 ==> du = 1/x dx and v = x and re-writing gives: uv - ∫ v du = (x)[ln(x)] - ∫ x (1/x dx) = x*ln(x) - ∫ 1 dx = x*ln(x) -...
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Source: http://answers.yahoo.com/question/index?qid=20091001055116AAQr4jv
