integral greatest integer of x between the limits 15 to 11

To find the integral of the greatest integer function, also known as the floor function, between the limits of 15 to 11, we first need to clarify what we’re calculating. The greatest integer function, denoted as ⌊x⌋, returns the largest integer less than or equal to x. In this case, we are interested in the integral of ⌊x⌋ from 11 to 15.

Setting Up the Integral

The integral we want to evaluate is:

∫₁₁¹⁵ ⌊x⌋ dx

Since the floor function is piecewise constant over intervals of integers, we can break this integral into segments where ⌊x⌋ remains constant.

Identifying the Intervals

• From 11 to 12: ⌊x⌋ = 11
• From 12 to 13: ⌊x⌋ = 12
• From 13 to 14: ⌊x⌋ = 13
• From 14 to 15: ⌊x⌋ = 14

Now we can express the integral as the sum of integrals over these intervals:

∫₁₁¹⁵ ⌊x⌋ dx = ∫₁₁¹² 11 dx + ∫₁₂¹³ 12 dx + ∫₁₃¹⁴ 13 dx + ∫₁₄¹⁵ 14 dx

Calculating Each Integral

Now, we can compute each of these integrals individually:

• ∫₁₁¹² 11 dx = 11(x) |₁₁¹² = 11(12 - 11) = 11
• ∫₁₂¹³ 12 dx = 12(x) |₁₂¹³ = 12(13 - 12) = 12
• ∫₁₃¹⁴ 13 dx = 13(x) |₁₃¹⁴ = 13(14 - 13) = 13
• ∫₁₄¹⁵ 14 dx = 14(x) |₁₄¹⁵ = 14(15 - 14) = 14

Summing the Results

Now, let's add up these results:

11 + 12 + 13 + 14 = 50

The Final Value

Thus, the integral of the greatest integer function ⌊x⌋ from 11 to 15 is equal to 50. This means that, when you consider the area under the curve of the greatest integer function across these limits, you get a total area of 50 square units.