Final answer:
To approximate the integral using the trapezoidal rule with four partitions, we calculate the width of each partition and apply the trapezoidal rule formula. After substituting the function values and calculating, we find that the approximation is 1.5625, nearest to answer option C 1.394.
Step-by-step explanation:
We need to approximate \(\int_{1}^{2} x^{1} dx\) using the trapezoidal rule with four equal partitions. The interval [1,2] is divided into four equal parts which means each part has a width of \(\frac{2-1}{4} = 0.25\). Let's call the partitions x_0, x_1, x_2, x_3, and x_4 where x_0 = 1 and x_4 = 2.
The trapezoidal rule is given by:
\(T = \frac{width}{2} \times (f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4))\)
Substituting the width and the function values, we get:
\(T = \frac{0.25}{2} \times (f(1) + 2f(1.25) + 2f(1.5) + 2f(1.75) + f(2))\)
\(T = 0.125 \times (1 + 2 \times 1.25 + 2 \times 1.5 + 2 \times 1.75 + 2)\)
\(T = 0.125 \times (1 + 2.5 + 3 + 3.5 + 2) = 0.125 \times 12.5\)
\(T = 1.5625\)
Therefore, the approximation of the integral using the trapezoidal rule is 1.5625, which is closest to the option C 1.394 when rounded to the nearest thousandths.