The value of the definite integral `int_0^1(x\ dx)/ (x^3+16)` lies in the interval `[a,b].` Then smallest such interval is

Options

A. `[0, (1)/(17)]`
B. `[0, 1]`
C. `[0, (1)/(27)]`
D. None of these

Answer

To solve the definite integral \(\int_0^1 \frac{x}{x^3 + 16} \, dx\) and find the interval \([a, b]\) that contains its value, we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int_0^1 \frac{x}{x^3 + 16} \, dx \]

### Step 2: Use a Substitution
To simplify the integral, we can use the substitution \(t = x^{3/2}\). Then, we differentiate to find \(dx\):
\[ x = t^{2/3} \quad \Rightarrow \quad dx = \frac{2}{3} t^{-1/3} \, dt \]
Now we need to change the limits of integration. When \(x = 0\), \(t = 0^{3/2} = 0\), and when \(x = 1\), \(t = 1^{3/2} = 1\).

### Step 3: Substitute into the Integral
Substituting \(x\) and \(dx\) into the integral gives:
\[ I = \int_0^1 \frac{t^{2/3}}{(t^{2/3})^3 + 16} \cdot \frac{2}{3} t^{-1/3} \, dt = \frac{2}{3} \int_0^1 \frac{t^{2/3 - 1/3}}{t^2 + 16} \, dt = \frac{2}{3} \int_0^1 \frac{t^{1/3}}{t^2 + 16} \, dt \]

### Step 4: Evaluate the Integral
Now we need to evaluate:
\[ I = \frac{2}{3} \int_0^1 \frac{t^{1/3}}{t^2 + 16} \, dt \]
This integral can be evaluated using numerical methods or further substitutions, but we can also estimate its value.

### Step 5: Estimate the Integral
To find bounds for the integral, we can analyze the function \(\frac{x}{x^3 + 16}\):
- At \(x = 0\), \(\frac{0}{0 + 16} = 0\).
- At \(x = 1\), \(\frac{1}{1 + 16} = \frac{1}{17}\).

Since \(\frac{x}{x^3 + 16}\) is continuous and positive on \([0, 1]\), we can find bounds for the integral:
\[ 0 < I < \int_0^1 \frac{1}{16} \, dx = \frac{1}{16} \]

### Step 6: Final Interval
Thus, we can conclude that:
\[ 0 < I < \frac{1}{16} \approx 0.0625 \]
This means the value of the definite integral lies in the interval \((0, 0.0625)\).

### Conclusion
The smallest interval \([a, b]\) that contains the value of the integral is:
\[ [0, 0.0625] \]