Question #186252

Improper Integral (Integrals with Infinite Limits)


∫dy/√(y-1) from 5 to ∞


Expert's answer

Solution:



∫1y−1 dxfrom5 to ∞\int \frac{1}{\sqrt{y-1}}\,dx \quad from \quad 5\ to \ \infty


can be written as


lim⁡a→∞∫5a1y−1 dy\lim_{a \to \infty} \int_{5}^a \frac{1}{\sqrt{y-1}}\,dy

y−1=t  ⟹   dy=dty→5;t→4 and y→∞;t→∞∴lim⁡a→∞∫4a1t dty-1 =t \implies\,dy=dt\\ y\rightarrow5; t\rightarrow4\ and\ y\rightarrow\infty; t\rightarrow\infty\\ \therefore \quad \lim_{a \to \infty} \int_{4}^a \frac{1}{\sqrt{t}}\,dt\\lim⁡a→∞[[t−12+1−12+1]a−[t−12+1−12+1]4]\lim_{a \to \infty} \lbrack \lbrack \frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\rbrack _{a}-\lbrack \frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\rbrack _{4}\rbrack

lim⁡a→∞[2a−24]lim⁡a→∞[2a−4]\lim_{a \to \infty} \lbrack 2\sqrt{a} -2\sqrt{4}\rbrack \\ \lim_{a \to \infty} \lbrack 2\sqrt{a} -4\rbrack\\

As a tends to infinite, entire expression tends to infinite.

Thus the given integral is divergent.

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