To approximate the definite integral using a series, we need to know the function and the interval of integration. Since you haven't provided this information, I am unable to give a specific answer. However, I can provide a general approach for using series to approximate integrals.
One commonly used series for approximating integrals is the Taylor series expansion. The Taylor series represents a function as an infinite sum of terms, which allows us to approximate the function within a certain range.
To approximate the definite integral, we can use the Taylor series expansion of the function and integrate each term of the series individually. This is known as term-by-term integration.
The accuracy of the approximation depends on the number of terms included in the series. Adding more terms increases the precision but also increases the computational complexity. Typically, we stop adding terms when the desired level of accuracy is achieved.
To provide a specific approximation, I would need the function and the interval of integration. If you can provide these details, I would be happy to help you with the series approximation of the definite integral, giving the answer correct to 3 decimal places.
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Use series to approximate the definite integral I. (Give your answer correct to 3 decimal places.) I = int_0^1 2 x cos\(x^2\)dx
Determine the convergence or divergence of the sequence with the given nth term. if the sequence converges, find its limit. (if the quantity diverges, enter diverges. ) an = 5 n 5 n 8
The limit of the sequence as n approaches infinity is 1. Since the sequence converges to a specific value (1).
To determine the convergence or divergence of the sequence with the given nth term, let's examine the expression:
an = 5n / (5n + 8)
As n approaches infinity, we can analyze the behavior of the sequence.
First, let's simplify the expression by dividing both the numerator and denominator by n:
an = (5n/n) / [(5n + 8)/n]
= 5 / (5 + 8/n)
As n approaches infinity, the term 8/n approaches zero since n is increasing without bound. Therefore, we have:
an ≈ 5/5
an ≈ 1
Hence, the limit of the sequence as n approaches infinity is 1.
Since the sequence converges to a specific value (1), we can conclude that the sequence converges.
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