Questions: Follow the steps below to compute the improper integral [ int2^infty frac3x^2 d x ] a. Use the Fundamental Theorem of Calculus to compute a related proper integral: [ int2^R frac3x^2 d x= ] b. Take the limit as R → ∞ of the result to compute the improper integral. Use ∞ if appropriate. [ int2^infty frac3x^2 d x=lim R rightarrow infty int2^R frac3x^2 d x= ] c. Is this improper integral convergent (finite) or divergent (not finite)? This improper integral converges. This improper integral diverges.

Solution Steps

To solve the given improper integral, we will follow these steps:

a. Use the Fundamental Theorem of Calculus to evaluate the definite integral from 2 to R of the function \(\frac{3}{x^2}\). This involves finding the antiderivative of the function and then evaluating it at the bounds 2 and R.

b. Compute the limit of the result from part (a) as \(R\) approaches infinity. This will determine the value of the improper integral.

c. Determine whether the improper integral is convergent or divergent based on the result from part (b).

To evaluate the definite integral \(\int_{2}^{R} \frac{3}{x^2} \, dx\), we first find the antiderivative of \(\frac{3}{x^2}\). The antiderivative is \(-\frac{3}{x}\). Applying the Fundamental Theorem of Calculus, we evaluate this antiderivative at the bounds 2 and \(R\):

\[ \int_{2}^{R} \frac{3}{x^2} \, dx = \left[-\frac{3}{x}\right]_{2}^{R} = -\frac{3}{R} + \frac{3}{2} \]

Next, we take the limit of the result from Step 1 as \(R\) approaches infinity:

\[ \lim_{R \to \infty} \left(-\frac{3}{R} + \frac{3}{2}\right) = \lim_{R \to \infty} \left(-\frac{3}{R}\right) + \frac{3}{2} = 0 + \frac{3}{2} = \frac{3}{2} \]

Since the limit exists and is finite, the improper integral \(\int_{2}^{\infty} \frac{3}{x^2} \, dx\) converges.

Final Answer