Answer:
R3 is not a right Riemann sum it is actually a notation for the midpoint Riemann sum. The right Riemann sum is denoted as Rn.
To approximate the integral \( \int_{1}^{4} x^{2} dx \) using R3 we need to divide the interval [1 4] into 3 subintervals of equal width.
The width of each subinterval is given by \( \Delta x = \frac{b-a}{n} = \frac{4-1}{3} = 1 \).
The right endpoints of the subintervals are: x1 = 2 x2 = 3 and x3 = 4.
Now we can evaluate the function \( f(x) = x^{2} \) at each of these right endpoints and sum up the areas of the corresponding rectangles.
R3 approximation:
\( R_{3} = f(2) \Delta x + f(3) \Delta x + f(4) \Delta x \)
\( R_{3} = 2^{2} \cdot 1 + 3^{2} \cdot 1 + 4^{2} \cdot 1 \)
\( R_{3} = 4 + 9 + 16 \)
\( R_{3} = 29 \)
Now let's answer your questions:
(1) The error of this approximation can be calculated by finding the difference between the value of the actual integral and the value of the R3 approximation:
\( \text{Error} = \int_{1}^{4} x^{2} dx - R_{3} \)
(2) The absolute error of this approximation is the absolute value of the error it gives us the size of the deviation:
\( \text{Absolute Error} = \left| \int_{1}^{4} x^{2} dx - R_{3} \right| \)
(3) The third question is incomplete and cannot be answered based on the given information.
Now to calculate the error and absolute error we need to evaluate the actual integral:
\( \int_{1}^{4} x^{2} dx = \frac{x^{3}}{3} \Big|_{1}^{4} \)
\( \int_{1}^{4} x^{2} dx = \frac{4^{3}}{3} - \frac{1^{3}}{3} \)
\( \int_{1}^{4} x^{2} dx = \frac{64}{3} - \frac{1}{3} \)
\( \int_{1}^{4} x^{2} dx = \frac{63}{3} \)
\( \int_{1}^{4} x^{2} dx = 21 \)
Now we can calculate the error and absolute error:
\( \text{Error} = \int_{1}^{4} x^{2} dx - R_{3} = 21 - 29 = -8 \)
\( \text{Absolute Error} = \left| \int_{1}^{4} x^{2} dx - R_{3} \right| = |-8| = 8 \)
Therefore the error of this approximation is -8 and the absolute error is 8.