The image you uploaded shows a problem involving the trapezium rule for numerical integration. The goal seems to be to approximate the integral of $\int_0^2 \sqrt{1 + x^3} \, dx$, correct to four decimal places, and calculate the percentage error when using the trapezium rule with two and four strips.

Key details:

• The exact value of the integral is given as 3.2413.
• You are asked to approximate this integral using the trapezium rule with:
  1. Two strips.
  2. Four strips.
• For each case, the percentage error of the approximation is required, correct to two decimal places.

Steps to solve the problem:

1. Trapezium Rule Formula: For an integral $\int_a^b f(x) \, dx$, the trapezium rule with $n$ strips is: $\text{Approximation} = \frac{h}{2} \left[ f(a) + 2 \sum_{i=1}^{n-1} f(x_i) + f(b) \right]$ where $h = \frac{b - a}{n}$ and $x_i = a + ih$ for $i = 1, 2, \dots, n-1$.

2. Apply the rule with 2 strips:

   • $a = 0$, $b = 2$, and $n = 2$, so $h = \frac{2 - 0}{2} = 1$.
   • The function $f(x) = \sqrt{1 + x^3}$.
   • Evaluate at $x = 0$, $x = 1$, and $x = 2$.
     • $f(0) = \sqrt{1 + 0^3} = 1$,
     • $f(1) = \sqrt{1 + 1^3} = \sqrt{2} \approx 1.4142$,
     • $f(2) = \sqrt{1 + 2^3} = \sqrt{9} = 3$.
   • Now, use the trapezium rule: [ \text{Approximation with 2 strips} = \frac{1}{2} \left[1 + 2(1.4142) + 3 \right] = \frac{1}{2} \left[1 + 2.8284 + 3 \right] = \frac{1}{2} (6.8284) = 3.4142. ]

3. Percentage error for 2 strips: The percentage error is given by: $\text{Percentage error} = \left| \frac{\text{Exact value} - \text{Approximate value}}{\text{Exact value}} \right| \times 100.$ Using the exact value $3.2413$ and the approximation $3.4142$, we get: $\text{Percentage error} = \left| \frac{3.2413 - 3.4142}{3.2413} \right| \times 100 = \left| \frac{-0.1729}{3.2413} \right| \times 100 \approx 5.33\%.$

4. Apply the rule with 4 strips:

   • For $n = 4$, $h = \frac{2 - 0}{4} = 0.5$.
   • Evaluate at $x = 0$, $x = 0.5$, $x = 1$, $x = 1.5$, and $x = 2$:
     • $f(0) = 1$,
     • $f(0.5) = \sqrt{1 + 0.5^3} \approx 1.1339$,
     • $f(1) = 1.4142$,
     • $f(1.5) = \sqrt{1 + 1.5^3} \approx 2.1213$,
     • $f(2) = 3$.
   • Using the trapezium rule: [ \text{Approximation with 4 strips} = \frac{0.5}{2} \left[1 + 2(1.1339 + 1.4142 + 2.1213) + 3 \right] = 0.25 \left[1 + 2(4.6694) + 3 \right] ] $= 0.25 \left[1 + 9.3388 + 3 \right] = 0.25 (13.3388) = 3.3347.$

5. Percentage error for 4 strips: Using the exact value $3.2413$ and the approximation $3.3347$, we get: $\text{Percentage error} = \left| \frac{3.2413 - 3.3347}{3.2413} \right| \times 100 = \left| \frac{-0.0934}{3.2413} \right| \times 100 \approx 2.88\%.$

Answers:

• Approximate value with 2 strips: $3.4142$, percentage error: $5.33\%$.
• Approximate value with 4 strips: $3.3347$, percentage error: $2.88\%$.

Would you like more detailed steps or have any questions?

Here are 5 related questions:

1. How does the number of strips affect the accuracy of the trapezium rule?
2. Can you use Simpson’s rule to approximate this integral and compare it with the trapezium rule?
3. What is the derivation of the error bound for the trapezium rule?
4. How would you calculate the exact integral using a symbolic method?
5. What happens if you increase the number of strips further (e.g., to 10 or 20)?

Tip: The accuracy of the trapezium rule improves as the number of strips increases, but for higher accuracy, using Simpson's rule or other higher-order methods can be more efficient.