y = 0.5x3 - 3x2 +4x + 12


What is the approximate value of the integral between x=0 and x=8 using the Trapezium rule, with a step size of 2?

To approximate the value of the integral using the Trapezium Rule, we follow these steps:

### Step 1: Function and interval
The given function is:
\[ y(x) = 0.5x^3 - 3x^2 + 4x + 12 \]

We are asked to find the integral of \( y(x) \) between \( x = 0 \) and \( x = 8 \), with a step size \( h = 2 \).

### Step 2: Trapezium Rule formula
The Trapezium Rule formula for approximating an integral is:
\[
\int_{a}^{b} y(x) \, dx \approx \frac{h}{2} \left[ y_0 + 2(y_1 + y_2 + \dots + y_{n-1}) + y_n \right]
\]
where:
- \( h = 2 \) (step size),
- \( y_0, y_1, \dots, y_n \) are the function values at \( x_0, x_1, \dots, x_n \),
- \( x_0 = 0 \), \( x_n = 8 \), and \( n = \frac{b - a}{h} = 4 \).

### Step 3: Calculate \( y(x) \) at each step
We evaluate the function \( y(x) = 0.5x^3 - 3x^2 + 4x + 12 \) at the points \( x = 0, 2, 4, 6, 8 \).

- \( y(0) = 0.5(0)^3 - 3(0)^2 + 4(0) + 12 = 12 \)
- \( y(2) = 0.5(2)^3 - 3(2)^2 + 4(2) + 12 = 0.5(8) - 3(4) + 8 + 12 = 4 - 12 + 8 + 12 = 12 \)
- \( y(4) = 0.5(4)^3 - 3(4)^2 + 4(4) + 12 = 0.5(64) - 3(16) + 16 + 12 = 32 - 48 + 16 + 12 = 12 \)
- \( y(6) = 0.5(6)^3 - 3(6)^2 + 4(6) + 12 = 0.5(216) - 3(36) + 24 + 12 = 108 - 108 + 24 + 12 = 36 \)
- \( y(8) = 0.5(8)^3 - 3(8)^2 + 4(8) + 12 = 0.5(512) - 3(64) + 32 + 12 = 256 - 192 + 32 + 12 = 108 \)

### Step 4: Apply the Trapezium Rule
Now, applying the Trapezium Rule:
\[
\int_{0}^{8} y(x) \, dx \approx \frac{2}{2} \left[ 12 + 2(12 + 12 + 36) + 108 \right]
\]
\[
= 1 \times \left[ 12 + 2(60) + 108 \right]
\]
\[
= 1 \times \left[ 12 + 120 + 108 \right] = 240
\]

### Approximate value of the integral
The approximate value of the integral is \( 240 \).

Would you like more details or have any questions?

Here are five related questions:
1. How does the accuracy of the Trapezium Rule change with different step sizes?
2. What are the other methods for approximating integrals?
3. How does the Trapezium Rule compare to Simpson's Rule in this example?
4. How can you determine if the function is increasing or decreasing from its derivative?
5. Can you apply the Trapezium Rule to functions with varying step sizes?

**Tip:** The smaller the step size \( h \), the more accurate the approximation from the Trapezium Rule.

What is the approximate value of the integral between x=0 and x=8 using the Trapezium rule, with a step size of 2 for the function y = 0.5x^3 - 3x^2 + 4x + 12?

Learn how to approximate the integral of y = 0.5x^3 - 3x^2 + 4x + 12 between x=0 and x=8 using the Trapezium Rule with a step size of 2. This solution covers key concepts in numerical integration and is suitable for students at the undergraduate level studying calculus.

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