Determine the average value of `y=2x+3` in the interval `0 le x le 1`.

Options

A. `1`
B. `5`
C. `3`
D. `4`

Explanation

To determine the average value of the function \( y = 2x + 3 \) over the interval \( 0 \leq x \leq 1 \), we can follow these steps:

### Step 1: Understand the formula for average value
The average value of a function \( y \) over an interval \([a, b]\) is given by the formula:

\[ \text{Average value} = \frac{1}{b-a} \int_a^b y \, dx \]

In this case, \( a = 0 \) and \( b = 1 \).

### Step 2: Set up the integral
Using the formula, we can express the average value of \( y \) as:

\[ \text{Average value} = \frac{1}{1-0} \int_0^1 (2x + 3) \, dx \]

This simplifies to:

\[ \text{Average value} = \int_0^1 (2x + 3) \, dx \]

### Step 3: Break down the integral
We can split the integral into two parts:

\[ \int_0^1 (2x + 3) \, dx = \int_0^1 2x \, dx + \int_0^1 3 \, dx \]

### Step 4: Calculate the first integral
Now, we calculate the first integral:

\[ \int_0^1 2x \, dx \]

Using the power rule for integration, we have:

\[ \int 2x \, dx = 2 \cdot \frac{x^2}{2} = x^2 \]

Evaluating this from 0 to 1:

\[ \left[ x^2 \right]_0^1 = 1^2 - 0^2 = 1 \]

### Step 5: Calculate the second integral
Next, we calculate the second integral:

\[ \int_0^1 3 \, dx \]

This is simply:

\[ 3 \cdot x \bigg|_0^1 = 3 \cdot (1 - 0) = 3 \]

### Step 6: Combine the results
Now, we combine the results of both integrals:

\[ \int_0^1 (2x + 3) \, dx = 1 + 3 = 4 \]

### Step 7: Final average value
Thus, the average value of \( y \) over the interval \( [0, 1] \) is:

\[ \text{Average value} = 4 \]

### Conclusion
The average value of \( y = 2x + 3 \) in the interval \( 0 \leq x \leq 1 \) is \( 4 \).

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