find the error in the trapezoidal approximation T10 for integral 1 to 2 11e^1/x dx

To find the error in the Trapezoidal approximation \( T_{10} \) for the integral:

\[
I = \int_1^2 11e^{\frac{1}{x}} dx
\]

we can use the error formula for the Trapezoidal Rule. The error \( E_T \) for the Trapezoidal approximation is given by:

\[
E_T = -\frac{(b - a)^3}{12n^2} f''(\xi)
\]

where:

- \( a = 1 \), \( b = 2 \) (the limits of integration),
- \( n = 10 \) (the number of subintervals),
- \( f(x) = 11e^{\frac{1}{x}} \) (the function being integrated),
- \( f''(\xi) \) is the second derivative of \( f(x) \) evaluated at some point \( \xi \in [1, 2] \).

### Step 1: Find the second derivative \( f''(x) \)
Let \( f(x) = 11e^{\frac{1}{x}} \). First, find the first derivative \( f'(x) \):

\[
f'(x) = 11 \cdot e^{\frac{1}{x}} \cdot \left( -\frac{1}{x^2} \right)
= -\frac{11e^{\frac{1}{x}}}{x^2}
\]

Now, compute the second derivative \( f''(x) \):

\[
f''(x) = \frac{d}{dx} \left( -\frac{11e^{\frac{1}{x}}}{x^2} \right)
= -\frac{11}{x^2} \cdot \frac{d}{dx} \left( e^{\frac{1}{x}} \right) + e^{\frac{1}{x}} \cdot \frac{d}{dx} \left( -\frac{1}{x^2} \right)
\]

Using the chain rule:

\[
f''(x) = -\frac{11}{x^2} \cdot e^{\frac{1}{x}} \cdot \left( -\frac{1}{x^2} \right) + e^{\frac{1}{x}} \cdot \frac{2}{x^3}
\]

Simplifying:

\[
f''(x) = \frac{11e^{\frac{1}{x}}}{x^4} + \frac{22e^{\frac{1}{x}}}{x^5}
= e^{\frac{1}{x}} \left( \frac{11}{x^4} + \frac{22}{x^5} \right)
\]

### Step 2: Estimate \( f''(\xi) \)
Since we don't know \( \xi \), we can use a reasonable approximation by finding the maximum value of \( f''(x) \) on the interval \( [1, 2] \). The second derivative is largest when \( x \) is smallest, i.e., at \( x = 1 \). Let's calculate \( f''(1) \):

\[
f''(1) = e^{\frac{1}{1}} \left( \frac{11}{1^4} + \frac{22}{1^5} \right)
= e^1 (11 + 22)
= 33e
\]

### Step 3: Compute the error
Now use the error formula with \( f''(\xi) \approx f''(1) = 33e \):

\[
E_T = -\frac{(2 - 1)^3}{12 \cdot 10^2} \cdot 33e
= -\frac{1}{12 \cdot 100} \cdot 33e
= -\frac{33e}{1200}
\]

Numerically, since \( e \approx 2.718 \), the error is approximately:

\[
E_T \approx -\frac{33 \cdot 2.718}{1200}
= -0.0748
\]

Thus, the error in the Trapezoidal approximation \( T_{10} \) is approximately \( -0.0748 \).

---

Would you like more details or have any questions?

Here are 5 related questions to expand on this:

1. How do you derive the error formula for the Trapezoidal Rule?
2. How does the number of subintervals affect the accuracy of the Trapezoidal approximation?
3. Can you compute the integral \( \int_1^2 11e^{\frac{1}{x}} dx \) using Simpson's Rule?
4. What is the difference between Trapezoidal and Midpoint approximations for the same integral?
5. How can the error estimation be improved by finding the maximum of \( f''(x) \) more precisely?

**Tip:** The error in the Trapezoidal Rule decreases as \( 1/n^2 \), so doubling the number of intervals significantly improves accuracy.

Explore the error estimation for the Trapezoidal approximation T10 in evaluating the integral of 11e^(1/x) from 1 to 2. This problem demonstrates the use of the Trapezoidal Rule error formula, derivative calculations, and numerical methods. Suitable for students in undergraduate calculus.

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