The smallest interval [a,b] such that
`int_0^(1) (1)/(sqrt(1+x^(4)))dx in [a,b]`, is

Options

A. `[(1)/(sqrt(2)),1]`
B. [0,1]
C. `[(1)/(2),1]`
D. `[(3)/(4),1]`

Answer

To solve the problem of finding the smallest interval \([a,b]\) such that

\[ \int_0^1 \frac{1}{\sqrt{1+x^4}} \, dx \in [a,b], \]

we can follow these steps:

### Step 1: Define the integral
Let
\[ I = \int_0^1 \frac{1}{\sqrt{1+x^4}} \, dx. \]

### Step 2: Analyze the function
The function \(\frac{1}{\sqrt{1+x^4}}\) is continuous and decreasing on the interval \([0, 1]\) since as \(x\) increases, \(x^4\) increases, making the denominator larger and the whole fraction smaller.

### Step 3: Find the bounds of the function
At \(x = 0\):
\[ f(0) = \frac{1}{\sqrt{1+0^4}} = 1. \]
At \(x = 1\):
\[ f(1) = \frac{1}{\sqrt{1+1^4}} = \frac{1}{\sqrt{2}}. \]

### Step 4: Establish the inequality for the integral
Since \(f(x)\) is decreasing, we can establish the following inequalities for the integral:
\[ \int_0^1 f(1) \, dx < I < \int_0^1 f(0) \, dx. \]
This simplifies to:
\[ \frac{1}{\sqrt{2}} < I < 1. \]

### Step 5: Calculate the bounds
Calculating the integrals gives:
\[ \int_0^1 f(1) \, dx = \frac{1}{\sqrt{2}} \cdot (1 - 0) = \frac{1}{\sqrt{2}}, \]
and
\[ \int_0^1 f(0) \, dx = 1 \cdot (1 - 0) = 1. \]

### Step 6: Conclusion
Thus, we have:
\[ \frac{1}{\sqrt{2}} < I < 1. \]
The smallest interval \([a, b]\) that contains \(I\) is:
\[ \left[\frac{1}{\sqrt{2}}, 1\right]. \]

### Final Answer
The smallest interval \([a,b]\) such that \(I \in [a,b]\) is:
\[ \left[\frac{1}{\sqrt{2}}, 1\right]. \]