Questions: Determine whether the improper integral ∫ from 1 to ∞ of 7/x^3 dx converges or diverges. If it converges, find its value. If it diverges, enter DNE. Integral: The value of the integral is:

Solution Steps

To determine whether the improper integral \(\int_{1}^{\infty} \frac{7}{x^{3}} \, dx\) converges or diverges, we can evaluate the limit of the definite integral as the upper bound approaches infinity. If the limit exists and is finite, the integral converges; otherwise, it diverges. Specifically, we will compute \(\lim_{b \to \infty} \int_{1}^{b} \frac{7}{x^{3}} \, dx\).

We start by finding the indefinite integral of the function \( f(x) = \frac{7}{x^3} \):

\[ \int \frac{7}{x^3} \, dx = -\frac{7}{2x^2} + C \]

Next, we evaluate the definite integral from \( 1 \) to \( b \):

\[ \int_{1}^{b} \frac{7}{x^3} \, dx = \left[-\frac{7}{2x^2}\right]_{1}^{b} = -\frac{7}{2b^2} + \frac{7}{2} \]

Now, we compute the limit of the definite integral as \( b \) approaches infinity:

\[ \lim_{b \to \infty} \left(-\frac{7}{2b^2} + \frac{7}{2}\right) = 0 + \frac{7}{2} = \frac{7}{2} \]

Since the limit exists and is finite, the improper integral converges.

Final Answer