QuestionLet $f(x)=\sqrt{x^{2}+x}$ for $1 \leq x \leq 7$. We wish to estimate $$\int_{1}^{7} f(x) d x$$ by the Trapezoidal Rule. a) Divide the domain of $f$ into 4 sub-intervals of equal length. Calculate their common length $\Delta x$ (exact value). 1.5 $\square$ b) Find the approximation of $\int_{1}^{7} f(x) d x$ that the Trapezoidal Rule produces with 4 sub-intervals. Give the answer with $\pm 0.0001$ precision. Number

Studdy Solution
Apply the Trapezoidal Rule to estimate the integral:
The Trapezoidal Rule formula for $n$ sub-intervals is:
$\int_{a}^{b} f(x) \, dx \approx \frac{\Delta x}{2} \left( f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n) \right)$
Where $x_0, x_1, \ldots, x_n$ are the endpoints of the sub-intervals.
For our problem: - $a = 1$, $b = 7$, $n = 4$ - $\Delta x = 1.5$ - The endpoints are $x_0 = 1$, $x_1 = 2.5$, $x_2 = 4$, $x_3 = 5.5$, $x_4 = 7$.
Calculate $f(x)$ at each endpoint:
$f(x_0) = f(1) = \sqrt{1^2 + 1} = \sqrt{2}$ $f(x_1) = f(2.5) = \sqrt{(2.5)^2 + 2.5} = \sqrt{6.25 + 2.5} = \sqrt{8.75}$ $f(x_2) = f(4) = \sqrt{4^2 + 4} = \sqrt{16 + 4} = \sqrt{20}$ $f(x_3) = f(5.5) = \sqrt{(5.5)^2 + 5.5} = \sqrt{30.25 + 5.5} = \sqrt{35.75}$ $f(x_4) = f(7) = \sqrt{7^2 + 7} = \sqrt{49 + 7} = \sqrt{56}$
Apply the Trapezoidal Rule:
$\int_{1}^{7} f(x) \, dx \approx \frac{1.5}{2} \left( \sqrt{2} + 2\sqrt{8.75} + 2\sqrt{20} + 2\sqrt{35.75} + \sqrt{56} \right)$
$= 0.75 \left( \sqrt{2} + 2\sqrt{8.75} + 2\sqrt{20} + 2\sqrt{35.75} + \sqrt{56} \right)$
Calculate the numerical approximation:
$\approx 0.75 \left( 1.4142 + 2 \times 2.9580 + 2 \times 4.4721 + 2 \times 5.9791 + 7.4833 \right)$
$\approx 0.75 \left( 1.4142 + 5.9160 + 8.9442 + 11.9582 + 7.4833 \right)$
$\approx 0.75 \times 35.7159$
$\approx 26.7869$
The approximation of $\int_{1}^{7} f(x) \, dx$ using the Trapezoidal Rule with 4 sub-intervals is approximately $\boxed{26.7869}$.