How do you use the direct comparison test for improper integrals?

Answer 1

If an improper integral diverges, then we probably do not want to wast time trying to find its value.

Let us assume that we already know:

#int_1^infty1/x dx=infty#

Let us look examine this uglier improper integral.

#int_1^infty{4x^2+5x+8}/{3x^3-x-1} dx#

By making the numerator smaller and the denominator bigger,

#{4x^2+5x+8}/{3x^3-x-1} ge {3x^2}/{3x^3}=1/x#

By Comparison Test, we may conclude that

#int_1^infty{4x^2+5x+8}/{3x^3-x-1} dx# diverges.

(Intuitively, if the smaller integral diverges, then the larger one has no chance to converge.)

I hope that this was helpful.

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Answer 2

The direct comparison test for improper integrals states that if (f(x)) and (g(x)) are continuous functions for (x \geq a), where (a) is a constant, and (0 \leq f(x) \leq g(x)) for all (x \geq a), then either both integrals (\int_{a}^{\infty} f(x) , dx) and (\int_{a}^{\infty} g(x) , dx) converge or both diverge.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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