Evaluate the integral as limit of sum: `int_(0)^(1) (e^(2x)-e^(x) +x) dx`

Answer

To evaluate the integral \( \int_{0}^{1} (e^{2x} - e^{x} + x) \, dx \) as a limit of sums, we will first compute the integral using the Fundamental Theorem of Calculus and then express it in terms of a limit of sums.

### Step-by-step Solution:

1. **Set up the integral**:
We want to evaluate the integral:
\[ I = \int_{0}^{1} (e^{2x} - e^{x} + x) \, dx \]

2. **Integrate each term separately**:
- The integral of \( e^{2x} \):
\[ \int e^{2x} \, dx = \frac{e^{2x}}{2} \]
- The integral of \( e^{x} \):
\[ \int e^{x} \, dx = e^{x} \]
- The integral of \( x \):
\[ \int x \, dx = \frac{x^2}{2} \]

3. **Combine the integrals**:
Therefore, we can write:
\[ I = \left[ \frac{e^{2x}}{2} - e^{x} + \frac{x^2}{2} \right]_{0}^{1} \]

4. **Evaluate at the upper limit (x = 1)**:
Substitute \( x = 1 \):
\[ I = \left( \frac{e^{2 \cdot 1}}{2} - e^{1} + \frac{1^2}{2} \right) = \left( \frac{e^{2}}{2} - e + \frac{1}{2} \right) \]

5. **Evaluate at the lower limit (x = 0)**:
Substitute \( x = 0 \):
\[ I = \left( \frac{e^{2 \cdot 0}}{2} - e^{0} + \frac{0^2}{2} \right) = \left( \frac{1}{2} - 1 + 0 \right) = -\frac{1}{2} \]

6. **Combine the results**:
Now, we combine the results from the upper and lower limits:
\[ I = \left( \frac{e^{2}}{2} - e + \frac{1}{2} \right) - \left( -\frac{1}{2} \right) \]
Simplifying this gives:
\[ I = \frac{e^{2}}{2} - e + \frac{1}{2} + \frac{1}{2} = \frac{e^{2}}{2} - e + 1 \]

### Final Result:
Thus, the value of the integral is:
\[ I = \frac{e^{2}}{2} - e + 1 \]