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### Problem:
The gradient function is given by:
\[
\frac{dy}{dx} = 10e^{2x} - 5
\]
It is also given that when \( x = 0 \), \( y = 8 \). We need to find the value of \( y \) when \( x = 1 \).

### Solution:

1. **Integrate the derivative to find \( y \):**
   \[
   y = \int (10e^{2x} - 5) dx
   \]

2. **Integrate each term separately:**
   \[
   y = 10 \int e^{2x} dx - \int 5 dx
   \]

   - The integral of \( e^{2x} \) is \( \frac{1}{2}e^{2x} \), so:
     \[
     10 \int e^{2x} dx = 10 \cdot \frac{1}{2}e^{2x} = 5e^{2x}
     \]
   - The integral of 5 is:
     \[
     \int 5 dx = 5x
     \]

   Therefore:
   \[
   y = 5e^{2x} - 5x + C
   \]

3. **Use the initial condition \( x = 0 \) and \( y = 8 \) to find \( C \):**
   \[
   8 = 5e^{0} - 5(0) + C
   \]
   \[
   8 = 5 + C \implies C = 3
   \]

   So the equation for \( y \) becomes:
   \[
   y = 5e^{2x} - 5x + 3
   \]

4. **Find \( y \) when \( x = 1 \):**
   \[
   y = 5e^{2(1)} - 5(1) + 3
   \]
   \[
   y = 5e^{2} - 5 + 3
   \]
   \[
   y = 5e^{2} - 2
   \]

5. **Approximate the final value:**
   Using \( e \approx 2.718 \):
   \[
   e^{2} \approx 7.389
   \]
   \[
   y = 5 \cdot 7.389 - 2
   \]
   \[
   y \approx 36.945 - 2 = 34.945
   \]

Therefore, the value of \( y \) when \( x = 1 \) is approximately \( 34.945 \).

Would you like any more details or have any questions about this solution?

### Related Questions:
1. What would happen if the initial condition was different, say \( y = 10 \) when \( x = 0 \)?
2. How do you interpret the gradient function \( \frac{dy}{dx} \) physically or geometrically?
3. Can you solve for \( x \) when \( y = 20 \) using the function we derived?
4. How would the solution change if the differential equation was \( \frac{dy}{dx} = 10e^{2x} - 7 \)?
5. Can you determine the limit of the function \( y \) as \( x \) approaches infinity?

### Tip:
When integrating functions involving exponential terms, always pay attention to the coefficients in the exponent. It can often simplify the integration process if handled correctly.

A gradient function is given by dy/dx = 10e^(2x) - 5. When x = 0, y = 8. Find the value of y when x = 1.

Learn how to find the value of y when x = 1, given the gradient function dy/dx = 10e^(2x) - 5 and the initial condition y = 8 when x = 0. This solution includes integration of exponential functions and application of the fundamental theorem of calculus. Suitable for advanced high school students.

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