determine whether the following integral converges or diverges.\n3\n1\n∫\n—\ndx\n0\n√(x^(1/4)+x)\nas x…

determine whether the following integral converges or diverges.\n3\n1\n∫\n—\ndx\n0\n√(x^(1/4)+x)\nas x approaches 0 from above, which of the following is true about the behavior of the integrand 1/√(x^(1/4)+x)?\na. 1/√(x^(1/4)+x) < 1/√x\nb. 1/√x < 1/√(x^(1/4)+x)\nc. 1/√(x^(1/4)+x) > 1/x\nd. 1/√(x^(1/4)+x) < √(1/x^(1/4))

determine whether the following integral converges or diverges.\n3\n1\n∫\n—\ndx\n0\n√(x^(1/4)+x)\nas x approaches 0 from above, which of the following is true about the behavior of the integrand 1/√(x^(1/4)+x)?\na. 1/√(x^(1/4)+x) < 1/√x\nb. 1/√x < 1/√(x^(1/4)+x)\nc. 1/√(x^(1/4)+x) > 1/x\nd. 1/√(x^(1/4)+x) < √(1/x^(1/4))

Answer

Explanation:

Step1: Analyze the integrand near x = 0

As (x\to0^{+}), we consider the dominant - term in the denominator (\sqrt{\sqrt{x^{\frac{1}{4}}+x}}). When (x) is very small and positive, (\sqrt{x^{\frac{1}{4}}+x}\approx\sqrt{\sqrt{x^{\frac{1}{4}}}}) (since for (x\to0^{+}), (\sqrt{x^{\frac{1}{4}}}\gg x)). So the integrand (f(x)=\frac{1}{\sqrt{\sqrt{x^{\frac{1}{4}}+x}}}\approx\frac{1}{\sqrt{\sqrt{x^{\frac{1}{4}}}}}=\frac{1}{x^{\frac{1}{8}}}). We know that for a non - negative function (f(x)) and an improper integral (\int_{a}^{b}f(x)dx) with a singularity at (x = a), we can use the comparison test. The integral (\int_{0}^{b}\frac{1}{x^{p}}dx) converges if (p<1) and diverges if (p\geq1). Now, we want to compare (\frac{1}{\sqrt{\sqrt{x^{\frac{1}{4}}+x}}}) with a known function. Since (\sqrt{\sqrt{x^{\frac{1}{4}}+x}}>\sqrt{x}) for (x\in(0,3)) (because (\sqrt{x^{\frac{1}{4}}+x}>x) when (x>0)), then (\frac{1}{\sqrt{\sqrt{x^{\frac{1}{4}}+x}}}<\frac{1}{\sqrt{x}}).

Answer:

A. (\frac{1}{\sqrt{\sqrt{x^{\frac{1}{4}}+x}}}<\frac{1}{\sqrt{x}})