Question

\(\int_0^2 \frac{x^{\frac{8}{3}}}{|x - 1|^{\frac{5}{2}}} \, dx =\)

• A. \(\frac{215}{63}\)
• B. \(\frac{216}{315}\)
• C. \(\frac{216}{189}\)
• D. \(\frac{210}{63}\)

Hint:

For improper integrals, split at points of discontinuity and check convergence by comparing with \(\int x^{-p} \, dx\). Use the Beta function for integrals involving powers of sine and cosine.

Answer 1

The Correct Option is D

The integral is improper at \( x = 1 \) due to the denominator \( |x - 1|^{\frac{5}{2}} \). Split the integral: \[ I = \int_0^1 \frac{x^{\frac{8}{3}}}{(1 - x)^{\frac{5}{2}}} \, dx + \int_1^2 \frac{x^{\frac{8}{3}}}{(x - 1)^{\frac{5}{2}}} \, dx \] First integral: \(\int_0^1 \frac{x^{\frac{8}{3}}}{(1 - x)^{\frac{5}{2}}} \, dx\). Substitute \( x = \sin^2 \theta \), \( dx = 2 \sin \theta \cos \theta \, d\theta \), limits from \( x = 0 \) (\(\theta = 0\)) to \( x = 1 \) (\(\theta = \frac{\pi}{2}\)): \[ x^{\frac{8}{3}} = (\sin^2 \theta)^{\frac{8}{3}} = \sin^{\frac{16}{3}} \theta, \quad 1 - x = 1 - \sin^2 \theta = \cos^2 \theta, \quad (1 - x)^{\frac{5}{2}} = (\cos^2 \theta)^{\frac{5}{2}} = \cos^5 \theta \] \[ \int_0^1 \frac{x^{\frac{8}{3}}}{(1 - x)^{\frac{5}{2}}} \, dx = \int_0^{\frac{\pi}{2}} \frac{\sin^{\frac{16}{3}} \theta}{\cos^5 \theta} \cdot 2 \sin \theta \cos \theta \, d\theta = 2 \int_0^{\frac{\pi}{2}} \sin^{\frac{19}{3}} \theta \cos^{-4} \theta \, d\theta \] Use the Beta function: For \(\int_0^{\frac{\pi}{2}} \sin^a \theta \cos^b \theta \, d\theta = \frac{1}{2} B\left( \frac{a + 1}{2}, \frac{b + 1}{2} \right)\), with \( a = \frac{19}{3} \), \( b = -4 \): \[ \frac{a + 1}{2} = \frac{\frac{19}{3} + 1}{2} = \frac{\frac{22}{3}}{2} = \frac{11}{3}, \quad \frac{b + 1}{2} = \frac{-4 + 1}{2} = -\frac{3}{2} \] The Beta function \( B(m, n) \) requires \( m, n>0 \), but \(\frac{b + 1}{2} = -\frac{3}{2}<0\), indicating divergence. Test convergence: \[ \int_0^1 (1 - x)^{-\frac{5}{2}} x^{\frac{8}{3}} \, dx \] Near \( x = 1 \), the integrand behaves as \((1 - x)^{-\frac{5}{2}}\). Since \(-\frac{5}{2}<-1\), the integral diverges at \( x = 1^- \). Second integral: \(\int_1^2 \frac{x^{\frac{8}{3}}}{(x - 1)^{\frac{5}{2}}} \, dx\). Substitute \( u = x - 1 \), \( x = u + 1 \), \( dx = du \), limits from \( x = 1 \) (\( u = 0 \)) to \( x = 2 \) (\( u = 1 \)): \[ \int_0^1 \frac{(u + 1)^{\frac{8}{3}}}{u^{\frac{5}{2}}} \, du \] Near \( u = 0 \), \((u + 1)^{\frac{8}{3}} \approx 1\), so the integrand is \(\approx u^{-\frac{5}{2}}\). Since \(-\frac{5}{2}<-1\), this diverges at \( u = 0^+ \). Since both integrals diverge, the original integral diverges. Thus, none of the options (\(\frac{215}{63} \approx 3.413\), \(\frac{216}{315} \approx 0.686\), \(\frac{216}{189} \approx 1.143\), \(\frac{210}{63} \approx 3.333\)) are correct, as they imply a finite value.