Answer :
To approximate the integral [tex]\(\int_0^6 f(x) \, dx\)[/tex] using a left-hand sum with [tex]\(n = 3\)[/tex] subdivisions, follow these steps:
1. Define the function: [tex]\(f(x) = x^2 - 3x\)[/tex].
2. Determine the interval and the number of subdivisions:
- Interval: [tex]\([0, 6]\)[/tex]
- Number of subdivisions: [tex]\(n = 3\)[/tex]
3. Calculate the width of each subinterval:
- The width, [tex]\(\Delta x\)[/tex], is given by:
[tex]\[ \Delta x = \frac{b - a}{n} = \frac{6 - 0}{3} = 2 \][/tex]
4. Identify the left endpoints of the subintervals:
- The interval [tex]\([0, 6]\)[/tex] is divided into 3 subintervals:
[tex]\([0, 2]\)[/tex], [tex]\([2, 4]\)[/tex], and [tex]\([4, 6]\)[/tex].
- The left endpoints of these subintervals are [tex]\(x_0 = 0\)[/tex], [tex]\(x_1 = 2\)[/tex], and [tex]\(x_2 = 4\)[/tex].
5. Evaluate the function at each left endpoint:
- [tex]\(f(0) = 0^2 - 3(0) = 0\)[/tex]
- [tex]\(f(2) = 2^2 - 3(2) = 4 - 6 = -2\)[/tex]
- [tex]\(f(4) = 4^2 - 3(4) = 16 - 12 = 4\)[/tex]
6. Calculate the left-hand sum:
- The left-hand sum (approximation of the integral) is:
[tex]\[ L \approx \sum_{i=0}^{n-1} f(x_i) \Delta x \][/tex]
- For our function and intervals, this becomes:
[tex]\[ L \approx f(0) \Delta x + f(2) \Delta x + f(4) \Delta x \][/tex]
- Substituting the values we found:
[tex]\[ L \approx [f(0) \cdot 2] + [f(2) \cdot 2] + [f(4) \cdot 2] \][/tex]
[tex]\[ L \approx [0 \cdot 2] + [-2 \cdot 2] + [4 \cdot 2] \][/tex]
[tex]\[ L \approx 0 + (-4) + 8 \][/tex]
[tex]\[ L \approx 4 \][/tex]
Therefore, the left-hand sum approximation of [tex]\(\int_0^6 f(x) \, dx\)[/tex] is [tex]\(4\)[/tex].
[tex]\[ \boxed{4} \][/tex]
1. Define the function: [tex]\(f(x) = x^2 - 3x\)[/tex].
2. Determine the interval and the number of subdivisions:
- Interval: [tex]\([0, 6]\)[/tex]
- Number of subdivisions: [tex]\(n = 3\)[/tex]
3. Calculate the width of each subinterval:
- The width, [tex]\(\Delta x\)[/tex], is given by:
[tex]\[ \Delta x = \frac{b - a}{n} = \frac{6 - 0}{3} = 2 \][/tex]
4. Identify the left endpoints of the subintervals:
- The interval [tex]\([0, 6]\)[/tex] is divided into 3 subintervals:
[tex]\([0, 2]\)[/tex], [tex]\([2, 4]\)[/tex], and [tex]\([4, 6]\)[/tex].
- The left endpoints of these subintervals are [tex]\(x_0 = 0\)[/tex], [tex]\(x_1 = 2\)[/tex], and [tex]\(x_2 = 4\)[/tex].
5. Evaluate the function at each left endpoint:
- [tex]\(f(0) = 0^2 - 3(0) = 0\)[/tex]
- [tex]\(f(2) = 2^2 - 3(2) = 4 - 6 = -2\)[/tex]
- [tex]\(f(4) = 4^2 - 3(4) = 16 - 12 = 4\)[/tex]
6. Calculate the left-hand sum:
- The left-hand sum (approximation of the integral) is:
[tex]\[ L \approx \sum_{i=0}^{n-1} f(x_i) \Delta x \][/tex]
- For our function and intervals, this becomes:
[tex]\[ L \approx f(0) \Delta x + f(2) \Delta x + f(4) \Delta x \][/tex]
- Substituting the values we found:
[tex]\[ L \approx [f(0) \cdot 2] + [f(2) \cdot 2] + [f(4) \cdot 2] \][/tex]
[tex]\[ L \approx [0 \cdot 2] + [-2 \cdot 2] + [4 \cdot 2] \][/tex]
[tex]\[ L \approx 0 + (-4) + 8 \][/tex]
[tex]\[ L \approx 4 \][/tex]
Therefore, the left-hand sum approximation of [tex]\(\int_0^6 f(x) \, dx\)[/tex] is [tex]\(4\)[/tex].
[tex]\[ \boxed{4} \][/tex]
