How would I use the Comparison Test in calculus to solve the integral #(cos(4x) +3) / (3x^3 + 1)# from 1 to infinity?
I'm not sure what you mean by "solve".
By comparison, we can see that the integral converges.
graph{(y-100((cos (4x)+3)/(3x^3+1)))(y-((400)/(3x^3+1)))=0 [0.33, 11.433, -0.75, 4.8]} (Both multiplied by 100 for clarity of relationship.)
So
graph{(y-100((cos (4x)+3)/(3x^3+1)))(y-((400)/(x^3)))=0 [0.2, 17.985, -1.124, 7.766]}
(Again multiplied by 100 for clarity.)
So
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To use the Comparison Test to determine the convergence of the integral (\int_{1}^{\infty} \frac{\cos(4x) + 3}{3x^3 + 1} , dx), we need to compare it with a known integral whose convergence or divergence is known.
First, note that (\frac{\cos(4x) + 3}{3x^3 + 1} \leq \frac{4}{3x^3}) for (x \geq 1) (since (\cos(4x) + 3 \leq 4) and (3x^3 + 1 \geq 3x^3)).
Now, consider the integral (\int_{1}^{\infty} \frac{4}{3x^3} , dx). This integral is a p-series with (p = 3), which converges since (p > 1).
Since (\frac{\cos(4x) + 3}{3x^3 + 1} \leq \frac{4}{3x^3}) for (x \geq 1) and the integral (\int_{1}^{\infty} \frac{4}{3x^3} , dx) converges, by the Comparison Test, the integral (\int_{1}^{\infty} \frac{\cos(4x) + 3}{3x^3 + 1} , dx) also converges.
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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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