Question

1. approximate the following integrals using trapezoidal rule and find a bound for the error:

(a) \\(\int_{1}^{1.5} x^2 \ln x \\, dx\\)
(b) \\(\int_{0}^{\pi} e^{3x} \sin 2x \\, dx\\)
(c) \\(\int_{-0.25}^{0.25} \cos^2 x \\, dx\\)

1. approximate the following integrals using simpsons rule and find a bound for the error:

(a) \\(\int_{0}^{1} x^2 e^{-x} \\, dx\\)
(b) \\(\int_{0.75}^{1.3} (\sin^2 x - 2x \sin x + 1) \\, dx\\)
(c) \\(\int_{-0.5}^{0} x \ln(x + 1) \\, dx\\)

1. given the function \\(f\\) at the following values:

Explanation:

🆕 New Concept Discovered: Trapezoidal Rule and Error Bound
Approximating integrals using linear segments and bounding the error

We will solve Question 1(a): Approximate the integral \( \int_{1}^{1.5} x^2 \ln x \, dx \) using the Trapezoidal rule and find a bound for the error.

For a single interval \([a, b]\) (where \( n = 1 \)), the Trapezoidal rule is:

$$ T = \frac{b - a}{2} [f(a) + f(b)] $$

The error bound for the single-interval Trapezoidal rule is given by:

$$ |E| \le \frac{(b - a)^3}{12} \max_{x \in [a, b]} |f''(x)| $$

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Step 1: Apply the Trapezoidal Rule

Identify the parameters from the integral \( \int_{1}^{1.5} x^2 \ln x \, dx \):

• \( f(x) = x^2 \ln x \)
• \( a = 1 \)
• \( b = 1.5 \)
• Step size \( h = b - a = 1.5 - 1 = 0.5 \)

Evaluate the function at the endpoints:

• \( f(1) = 1^2 \ln(1) = 0 \)
• \( f(1.5) = (1.5)^2 \ln(1.5) = 2.25 \times 0.4054651 = 0.9122965 \)

Apply the formula:

$$ T = \frac{0.5}{2} [f(1) + f(1.5)] $$

$$ T = 0.25 \times [0 + 0.9122965] \approx 0.228074 $$

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Step 2: Find the Second Derivative of \( f(x) \)

To calculate the error bound, we need the second derivative of \( f(x) = x^2 \ln x \).

First derivative \( f'(x) \) using the product rule:

$$ f'(x) = \frac{d}{dx}(x^2) \cdot \ln x + x^2 \cdot \frac{d}{dx}(\ln x) $$

$$ f'(x) = 2x \ln x + x^2 \cdot \frac{1}{x} = 2x \ln x + x $$

Second derivative \( f''(x) \):

$$ f''(x) = \frac{d}{dx}(2x \ln x + x) $$

$$ f''(x) = 2 \ln x + 2x \cdot \frac{1}{x} + 1 $$

$$ f''(x) = 2 \ln x + 2 + 1 = 2 \ln x + 3 $$

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Step 3: Find the Maximum of \( |f''(x)| \) on \( [1, 1.5] \)

Since \( \ln x \) is a strictly increasing function for \( x > 0 \), the second derivative \( f''(x) = 2 \ln x + 3 \) is also strictly increasing on the interval \( [1, 1.5] \).

Thus, the maximum value of \( |f''(x)| \) on this interval occurs at the right endpoint, \( x = 1.5 \):

$$ \max_{x \in [1, 1.5]} |2 \ln x + 3| = 2 \ln(1.5) + 3 $$

$$ 2(0.4054651) + 3 = 3.810930 $$

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Step 4: Calculate the Error Bound

Substitute the values into the error bound formula:

$$ |E| \le \frac{(b - a)^3}{12} \max_{x \in [a, b]} |f''(x)| $$

$$ |E| \le \frac{(0.5)^3}{12} \times 3.810930 $$

$$ |E| \le \frac{0.125}{12} \times 3.810930 $$

$$ |E| \le 0.0104167 \times 3.810930 \approx 0.039697 $$

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Answer:

• Trapezoidal Approximation: \( \approx 0.228074 \)
• Error Bound: \( \le 0.039697 \)