Answer :
To determine if the integral
[tex]\[ \int_0^1 \frac{1}{x^2+\sqrt{x}} \, dx \][/tex]
converges without actually evaluating it, we can use the Direct Comparison Test. Here’s a detailed, step-by-step solution:
### Step 1: Identify and analyze the integrand
Given the function inside the integral is:
[tex]\[ f(x) = \frac{1}{x^2 + \sqrt{x}} \][/tex]
This function is continuous for [tex]\( x \)[/tex] in the interval [tex]\([0, 1]\)[/tex] except at [tex]\( x = 0 \)[/tex]. To apply the Direct Comparison Test, we seek a simpler function [tex]\(g(x)\)[/tex] such that [tex]\(f(x) \leq g(x)\)[/tex] and [tex]\(g(x)\)[/tex] is easier to integrate and analyze over the interval [tex]\([0, 1]\)[/tex].
### Step 2: Identify a suitable comparison function
We observe the behavior of [tex]\(x^2\)[/tex] and [tex]\(\sqrt{x}\)[/tex] on the interval [tex]\([0, 1]\)[/tex]:
1. For [tex]\( 0 \leq x < 1 \)[/tex], [tex]\(x^2\)[/tex] is always non-negative and thus [tex]\(\sqrt{x}\)[/tex] provides the dominant term in the denominator near [tex]\(x = 0\)[/tex].
2. We need a function [tex]\(g(x)\)[/tex] that is greater than or equal to [tex]\(\frac{1}{x^2 + \sqrt{x}}\)[/tex].
Let’s consider:
[tex]\[ g(x) = \frac{1}{\sqrt{x}} \][/tex]
since [tex]\(x^2 \geq 0\)[/tex] for [tex]\(x \in [0, 1]\)[/tex], it follows that:
[tex]\[ x^2 + \sqrt{x} \geq \sqrt{x} \][/tex]
Therefore,
[tex]\[ \frac{1}{x^2 + \sqrt{x}} \leq \frac{1}{\sqrt{x}} \][/tex]
for [tex]\(0 < x \leq 1\)[/tex].
### Step 3: Examine the comparison integral
We now evaluate whether the comparison integral converges:
[tex]\[ \int_0^1 \frac{1}{\sqrt{x}} \, dx \][/tex]
To integrate [tex]\(\frac{1}{\sqrt{x}}\)[/tex]:
[tex]\[ \int_0^1 \frac{1}{\sqrt{x}} \, dx = \int_0^1 x^{-\frac{1}{2}} \, dx \][/tex]
This simplifies to:
[tex]\[ \left[ 2x^{\frac{1}{2}} \right]_0^1 = 2 \left[ x^{\frac{1}{2}} \right]_0^1 = 2 \left( 1^{\frac{1}{2}} - 0^{\frac{1}{2}} \right) = 2(1) = 2 \][/tex]
Since the integral [tex]\(\int_0^1 \frac{1}{\sqrt{x}} \, dx\)[/tex] converges to a finite value, it provides an upper bound for [tex]\(\frac{1}{x^2 + \sqrt{x}}\)[/tex].
### Step 4: Apply the Direct Comparison Test
By the Direct Comparison Test:
- Since [tex]\( \frac{1}{x^2 + \sqrt{x}} \leq \frac{1}{\sqrt{x}} \)[/tex] for [tex]\(0 < x \leq 1\)[/tex], and
- The integral [tex]\(\int_0^1 \frac{1}{\sqrt{x}} \, dx\)[/tex] converges,
we can conclude that:
[tex]\[ \int_0^1 \frac{1}{x^2 + \sqrt{x}} \, dx \][/tex]
also converges.
Thus, based on the Direct Comparison Test, we have shown that the integral [tex]\(\int_0^1 \frac{1}{x^2 + \sqrt{x}} \, dx\)[/tex] converges.
[tex]\[ \int_0^1 \frac{1}{x^2+\sqrt{x}} \, dx \][/tex]
converges without actually evaluating it, we can use the Direct Comparison Test. Here’s a detailed, step-by-step solution:
### Step 1: Identify and analyze the integrand
Given the function inside the integral is:
[tex]\[ f(x) = \frac{1}{x^2 + \sqrt{x}} \][/tex]
This function is continuous for [tex]\( x \)[/tex] in the interval [tex]\([0, 1]\)[/tex] except at [tex]\( x = 0 \)[/tex]. To apply the Direct Comparison Test, we seek a simpler function [tex]\(g(x)\)[/tex] such that [tex]\(f(x) \leq g(x)\)[/tex] and [tex]\(g(x)\)[/tex] is easier to integrate and analyze over the interval [tex]\([0, 1]\)[/tex].
### Step 2: Identify a suitable comparison function
We observe the behavior of [tex]\(x^2\)[/tex] and [tex]\(\sqrt{x}\)[/tex] on the interval [tex]\([0, 1]\)[/tex]:
1. For [tex]\( 0 \leq x < 1 \)[/tex], [tex]\(x^2\)[/tex] is always non-negative and thus [tex]\(\sqrt{x}\)[/tex] provides the dominant term in the denominator near [tex]\(x = 0\)[/tex].
2. We need a function [tex]\(g(x)\)[/tex] that is greater than or equal to [tex]\(\frac{1}{x^2 + \sqrt{x}}\)[/tex].
Let’s consider:
[tex]\[ g(x) = \frac{1}{\sqrt{x}} \][/tex]
since [tex]\(x^2 \geq 0\)[/tex] for [tex]\(x \in [0, 1]\)[/tex], it follows that:
[tex]\[ x^2 + \sqrt{x} \geq \sqrt{x} \][/tex]
Therefore,
[tex]\[ \frac{1}{x^2 + \sqrt{x}} \leq \frac{1}{\sqrt{x}} \][/tex]
for [tex]\(0 < x \leq 1\)[/tex].
### Step 3: Examine the comparison integral
We now evaluate whether the comparison integral converges:
[tex]\[ \int_0^1 \frac{1}{\sqrt{x}} \, dx \][/tex]
To integrate [tex]\(\frac{1}{\sqrt{x}}\)[/tex]:
[tex]\[ \int_0^1 \frac{1}{\sqrt{x}} \, dx = \int_0^1 x^{-\frac{1}{2}} \, dx \][/tex]
This simplifies to:
[tex]\[ \left[ 2x^{\frac{1}{2}} \right]_0^1 = 2 \left[ x^{\frac{1}{2}} \right]_0^1 = 2 \left( 1^{\frac{1}{2}} - 0^{\frac{1}{2}} \right) = 2(1) = 2 \][/tex]
Since the integral [tex]\(\int_0^1 \frac{1}{\sqrt{x}} \, dx\)[/tex] converges to a finite value, it provides an upper bound for [tex]\(\frac{1}{x^2 + \sqrt{x}}\)[/tex].
### Step 4: Apply the Direct Comparison Test
By the Direct Comparison Test:
- Since [tex]\( \frac{1}{x^2 + \sqrt{x}} \leq \frac{1}{\sqrt{x}} \)[/tex] for [tex]\(0 < x \leq 1\)[/tex], and
- The integral [tex]\(\int_0^1 \frac{1}{\sqrt{x}} \, dx\)[/tex] converges,
we can conclude that:
[tex]\[ \int_0^1 \frac{1}{x^2 + \sqrt{x}} \, dx \][/tex]
also converges.
Thus, based on the Direct Comparison Test, we have shown that the integral [tex]\(\int_0^1 \frac{1}{x^2 + \sqrt{x}} \, dx\)[/tex] converges.
