Evaluate the integral below or show that the integral is divergent
5
∫ (3x-4)/(x^2-16) dx
0
Please help me... Thank you...
5
∫ (3x-4)/(x^2-16) dx
0
Please help me... Thank you...
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By Partial Fractions, you can write:
(3x - 4)/(x^2 - 16) = 2/(x + 4) + 1/(x - 4).
Since an infinite discontinuity exists at x = 4, split the integral up at 4 to give:
∫ (3x - 4)/(x^2 - 16) dx (from x=0 to 5)
= ∫ (3x - 4)/(x^2 - 16) dx (from x=0 to 4) + ∫ (3x - 4)/(x^2 - 16) dx (from x=4 to 5).
By evaluating the first integral:
∫ (3x - 4)/(x^2 - 16) dx (from x=0 to 4)
= ∫ [2/(x + 4) + 1/(x - 4)] dx (from x=0 to 4), by partial fractions
= (2ln|x + 4| + ln|x - 4|) (evaluated from x=0 to 4), by integrating
= 2ln|(x + 4)/(x - 4)| (evaluated from x=0 to 4)
= 2 * lim (n-->4-) ln|(n + 4)/(n - 4)|
= -infinity.
Since this integral is divergent, the integral in question is divergent.
Just a note: just because there is an infinite discontinuity doesn't mean that the integral diverges. For example:
∫ 1/√dx dx (from x=0 to 2) is convergent.
I hope this helps!
(3x - 4)/(x^2 - 16) = 2/(x + 4) + 1/(x - 4).
Since an infinite discontinuity exists at x = 4, split the integral up at 4 to give:
∫ (3x - 4)/(x^2 - 16) dx (from x=0 to 5)
= ∫ (3x - 4)/(x^2 - 16) dx (from x=0 to 4) + ∫ (3x - 4)/(x^2 - 16) dx (from x=4 to 5).
By evaluating the first integral:
∫ (3x - 4)/(x^2 - 16) dx (from x=0 to 4)
= ∫ [2/(x + 4) + 1/(x - 4)] dx (from x=0 to 4), by partial fractions
= (2ln|x + 4| + ln|x - 4|) (evaluated from x=0 to 4), by integrating
= 2ln|(x + 4)/(x - 4)| (evaluated from x=0 to 4)
= 2 * lim (n-->4-) ln|(n + 4)/(n - 4)|
= -infinity.
Since this integral is divergent, the integral in question is divergent.
Just a note: just because there is an infinite discontinuity doesn't mean that the integral diverges. For example:
∫ 1/√dx dx (from x=0 to 2) is convergent.
I hope this helps!
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The integral does not converge. To see why this is so, note that the denominator under the integral:
x^2 - 16
is a difference of two squares, so:
x^2 - 16 = (x + 4)(x - 4)
and since it is in the denominator of the function you want to integrate, the function has a vertical asymptote at x = 4 and x = -4. Since you are integrating from 0 to 5, you will pass over one of these asymptotes, making it an improper integral. Thus, you will have to break it into pieces, and rewrite the pieces as limits. Hopefully this is enough of a hint that you can finish it on your own by showing that at least one of those limits goes to infinity.
Good luck!
x^2 - 16
is a difference of two squares, so:
x^2 - 16 = (x + 4)(x - 4)
and since it is in the denominator of the function you want to integrate, the function has a vertical asymptote at x = 4 and x = -4. Since you are integrating from 0 to 5, you will pass over one of these asymptotes, making it an improper integral. Thus, you will have to break it into pieces, and rewrite the pieces as limits. Hopefully this is enough of a hint that you can finish it on your own by showing that at least one of those limits goes to infinity.
Good luck!