To evaluate the Riemann sum \( R_6 \) for the function \( f(x) = 4x^2 - 3x \) on the interval \( [0, 3] \) using \( n = 6 \) subintervals and right endpoints, we will perform the following steps:
-
Determine the width of each subinterval:
The width of each subinterval \( \Delta x \) is given by: \[ \Delta x = \frac{b - a}{n} = \frac{3 - 0}{6} = \frac{3}{6} = 0.5. \]
-
Identify the right endpoints:
The right endpoints \( x_i \) can be calculated as:
- \( x_1 = 0 + 0.5 = 0.5 \)
- \( x_2 = 0 + 1 \times 0.5 = 1.0 \)
- \( x_3 = 0 + 2 \times 0.5 = 1.5 \)
- \( x_4 = 0 + 3 \times 0.5 = 2.0 \)
- \( x_5 = 0 + 4 \times 0.5 = 2.5 \)
- \( x_6 = 0 + 5 \times 0.5 = 3.0 \)
Thus, the right endpoints are: \( 0.5, 1.0, 1.5, 2.0, 2.5, 3.0 \).
-
Calculate the function values at each endpoint:
- \( f(0.5) = 4(0.5)^2 - 3(0.5) = 4(0.25) - 1.5 = 1 - 1.5 = -0.5 \)
- \( f(1.0) = 4(1.0)^2 - 3(1.0) = 4 - 3 = 1 \)
- \( f(1.5) = 4(1.5)^2 - 3(1.5) = 4(2.25) - 4.5 = 9 - 4.5 = 4.5 \)
- \( f(2.0) = 4(2.0)^2 - 3(2.0) = 4(4) - 6 = 16 - 6 = 10 \)
- \( f(2.5) = 4(2.5)^2 - 3(2.5) = 4(6.25) - 7.5 = 25 - 7.5 = 17.5 \)
- \( f(3.0) = 4(3.0)^2 - 3(3.0) = 4(9) - 9 = 36 - 9 = 27 \)
-
Set up the Riemann sum:
Now, we'll calculate the Riemann sum \( R_6 \): \[ R_6 = \Delta x \left( f(x_1) + f(x_2) + f(x_3) + f(x_4) + f(x_5) + f(x_6) \right) \] Plugging in the values: \[ R_6 = 0.5 \left( f(0.5) + f(1.0) + f(1.5) + f(2.0) + f(2.5) + f(3.0) \right) \] Substituting the function values: \[ R_6 = 0.5 \left( -0.5 + 1 + 4.5 + 10 + 17.5 + 27 \right) \] Simplifying the sum: \[ R_6 = 0.5 \left( 60.5 \right) = 30.25. \]
Thus, the Riemann sum \( R_6 \) is \[ \boxed{30.25}. \]
1 answer