Determine whether the improper integral diverges or converges. \int_1^(\infty ) (8)/(\root(8)(x))dx Evaluate the integral if it converges. (If the quantity diverges, enter DIVERGES.)
The Correct Answer and Explanation is:
We are given the improper integral: I=∫1∞8×8 dxI = \int_1^{\infty} \frac{8}{\sqrt[8]{x}} \, dxI=∫1∞8x8dx
This is an improper integral because the upper limit of integration is infinity. To determine whether the integral converges or diverges, we must first express the integrand in a simpler form and evaluate the integral.
Step 1: Simplify the integrand
Recall that x8=x1/8\sqrt[8]{x} = x^{1/8}8x=x1/8. Therefore, the integrand becomes: 8×8=8x−1/8\frac{8}{\sqrt[8]{x}} = 8 x^{-1/8}8x8=8x−1/8
Thus, the integral becomes: I=∫1∞8x−1/8 dxI = \int_1^{\infty} 8 x^{-1/8} \, dxI=∫1∞8x−1/8dx
Step 2: Find the antiderivative
Now, we can evaluate the indefinite integral. The general rule for integrating xnx^nxn (where n≠−1n \neq -1n=−1) is: ∫xn dx=xn+1n+1+C\int x^n \, dx = \frac{x^{n+1}}{n+1} + C∫xndx=n+1xn+1+C
For our integral, the exponent is −1/8-1/8−1/8. Applying the rule, we get: ∫x−1/8 dx=x7/87/8=87×7/8\int x^{-1/8} \, dx = \frac{x^{7/8}}{7/8} = \frac{8}{7} x^{7/8}∫x−1/8dx=7/8×7/8=78x7/8
Thus, the antiderivative of 8x−1/88 x^{-1/8}8x−1/8 is: 8⋅87×7/8=647×7/88 \cdot \frac{8}{7} x^{7/8} = \frac{64}{7} x^{7/8}8⋅78x7/8=764x7/8
Step 3: Evaluate the improper integral
Now, we evaluate the improper integral from 1 to infinity. We compute the limit of the integral as the upper limit approaches infinity: I=limb→∞∫1b8x−1/8 dx=limb→∞[647×7/8]1bI = \lim_{b \to \infty} \int_1^b 8 x^{-1/8} \, dx = \lim_{b \to \infty} \left[ \frac{64}{7} x^{7/8} \right]_1^bI=b→∞lim∫1b8x−1/8dx=b→∞lim[764x7/8]1b
Substituting the limits of integration: I=limb→∞(647b7/8−647⋅17/8)I = \lim_{b \to \infty} \left( \frac{64}{7} b^{7/8} – \frac{64}{7} \cdot 1^{7/8} \right)I=b→∞lim(764b7/8−764⋅17/8) I=limb→∞647(b7/8−1)I = \lim_{b \to \infty} \frac{64}{7} \left( b^{7/8} – 1 \right)I=b→∞lim764(b7/8−1)
As b→∞b \to \inftyb→∞, b7/8→∞b^{7/8} \to \inftyb7/8→∞. Therefore, the expression diverges to infinity.
Conclusion
Since the limit diverges, the improper integral diverges. Thus, the correct answer is: DIVERGES\boxed{DIVERGES}DIVERGES
