To determine the average value of the function \( y = 2x + 3 \) over the interval \( [0, 1] \), we can follow these steps:
### Step 1: Understand the formula for average value
The average value of a function \( f(x) \) over the interval \( [a, b] \) is given by the formula:
\[
\text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx
\]
In our case, \( f(x) = 2x + 3 \), \( a = 0 \), and \( b = 1 \).
### Step 2: Set up the integral
Using the formula, we can set up the integral for our function:
\[
\text{Average value} = \frac{1}{1-0} \int_{0}^{1} (2x + 3) \, dx
\]
This simplifies to:
\[
\text{Average value} = \int_{0}^{1} (2x + 3) \, dx
\]
### Step 3: Calculate the integral
Now, we will calculate the integral:
\[
\int (2x + 3) \, dx = \int 2x \, dx + \int 3 \, dx
\]
Calculating each part:
- The integral of \( 2x \) is \( x^2 \).
- The integral of \( 3 \) is \( 3x \).
Thus, we have:
\[
\int (2x + 3) \, dx = x^2 + 3x + C
\]
### Step 4: Evaluate the definite integral from 0 to 1
Now we evaluate the definite integral from 0 to 1:
\[
\int_{0}^{1} (2x + 3) \, dx = \left[ x^2 + 3x \right]_{0}^{1}
\]
Calculating at the upper limit (1):
\[
1^2 + 3(1) = 1 + 3 = 4
\]
Calculating at the lower limit (0):
\[
0^2 + 3(0) = 0 + 0 = 0
\]
Now, subtract the lower limit from the upper limit:
\[
4 - 0 = 4
\]
### Step 5: Conclusion
Thus, the average value of \( y = 2x + 3 \) over the interval \( [0, 1] \) is:
\[
\text{Average value} = 4
\]
### Final Answer
The average value of \( y = 2x + 3 \) in the interval \( [0, 1] \) is \( 4 \).
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