Approximate the area under the curve y = x^3 + 2 from x = 2 to x = 5 using a left sum with 6 subdivisions.
The area is approximately=
just answer and explanation please and thanks.
again, each rectangle has width 1/2, so the area is
1/2 (f(2) + f(5/2) + ... + f(9/2))
To approximate the area under the curve using a left sum with 6 subdivisions, we divide the interval [2, 5] into 6 equal subintervals. The width of each subinterval, Δx, is found by dividing the length of the interval by the number of subdivisions:
Δx = (5 - 2) / 6
= 3 / 6
= 0.5
Next, we calculate the left endpoint of each subinterval, starting from x = 2 and incrementing by Δx:
x1 = 2
x2 = 2 + 0.5
= 2.5
x3 = 2.5 + 0.5
= 3
x4 = 3 + 0.5
= 3.5
x5 = 3.5 + 0.5
= 4
x6 = 4 + 0.5
= 4.5
Then, we evaluate the function at each left endpoint to find the corresponding y-values:
y1 = (x1)^3 + 2
= (2)^3 + 2
= 8 + 2
= 10
y2 = (x2)^3 + 2
= (2.5)^3 + 2
= 15.625 + 2
= 17.625
y3 = (x3)^3 + 2
= (3)^3 + 2
= 27 + 2
= 29
y4 = (x4)^3 + 2
= (3.5)^3 + 2
= 42.875 + 2
= 44.875
y5 = (x5)^3 + 2
= (4)^3 + 2
= 64 + 2
= 66
y6 = (x6)^3 + 2
= (4.5)^3 + 2
= 91.125 + 2
= 93.125
Finally, we calculate the sum of the areas of the rectangles formed by each subinterval by multiplying the width of each subinterval by its corresponding height and adding them together:
Area ≈ (Δx) * (y1 + y2 + y3 + y4 + y5 + y6)
= 0.5 * (10 + 17.625 + 29 + 44.875 + 66 + 93.125)
= 0.5 * 260.625
= 130.3125
Therefore, the area under the curve y = x^3 + 2 from x = 2 to x = 5, approximated using a left sum with 6 subdivisions, is approximately 130.3125 square units.
To approximate the area under the curve using a left sum, we first need to divide the interval [2, 5] into smaller subdivisions. In this case, we have 6 subdivisions, so the width of each subdivision, or Δx, can be calculated by taking the difference between the endpoints and dividing it by the number of subdivisions:
Δx = (5 - 2) / 6 = 3 / 6 = 0.5
Next, we need to evaluate the function at the left endpoint of each subdivision and multiply it by the width (Δx). The left endpoint of the first subdivision is x = 2. For subsequent subdivisions, we add Δx to the previous left endpoint. So, the left endpoints for the 6 subdivisions are:
x1 = 2
x2 = 2 + Δx
x3 = 2 + 2Δx
x4 = 2 + 3Δx
x5 = 2 + 4Δx
x6 = 2 + 5Δx
Since the function is y = x^3 + 2, we can substitute the values of x into the function to find the corresponding y-values.
Once we have the y-values, we calculate the area by summing up the products of each y-value and the width Δx:
Area ≈ (y1 * Δx) + (y2 * Δx) + (y3 * Δx) + (y4 * Δx) + (y5 * Δx) + (y6 * Δx)
Finally, we compute the approximation by plugging in the values:
Area ≈ [(2^3 + 2) * 0.5] + [(2.5^3 + 2) * 0.5] + [(3^3 + 2) * 0.5] + [(3.5^3 + 2) * 0.5] + [(4^3 + 2) * 0.5] + [(4.5^3 + 2) * 0.5]