Find the following definite integrals. $\int_{0}^{1.5} \lfloor x \rfloor dx$ $\int_{0}^{2.5} \lfloor x \rfloor dx$
Understand the Problem
The question asks to find the definite integral of the greatest integer function (floor function) from 0 to 1.5 and from 0 to 2.5.
Answer
$ \int_{0}^{1.5} \lfloor x \rfloor dx = 0.5 $ $ \int_{0}^{2.5} \lfloor x \rfloor dx = 2 $
Answer for screen readers
$ \int_{0}^{1.5} \lfloor x \rfloor dx = 0.5 $
$ \int_{0}^{2.5} \lfloor x \rfloor dx = 2 $
Steps to Solve
- Evaluate the first integral by splitting the interval
Since the floor function $\lfloor x \rfloor$ has different values over different intervals, we split the integral at integer points within the integration limits.
$$ \int_{0}^{1.5} \lfloor x \rfloor dx = \int_{0}^{1} \lfloor x \rfloor dx + \int_{1}^{1.5} \lfloor x \rfloor dx $$
- Evaluate the floor function within each integral
For $0 \leq x < 1$, $\lfloor x \rfloor = 0$. For $1 \leq x < 1.5$, $\lfloor x \rfloor = 1$.
- Substitute the values of floor function and evaluate the integrals
$$ \int_{0}^{1} 0 , dx + \int_{1}^{1.5} 1 , dx = 0 + [x]_{1}^{1.5} = 1.5 - 1 = 0.5 $$
- Evaluate the second integral by splitting the interval
Similarly, for the second integral:
$$ \int_{0}^{2.5} \lfloor x \rfloor dx = \int_{0}^{1} \lfloor x \rfloor dx + \int_{1}^{2} \lfloor x \rfloor dx + \int_{2}^{2.5} \lfloor x \rfloor dx $$
- Evaluate the floor function within each integral
For $0 \leq x < 1$, $\lfloor x \rfloor = 0$. For $1 \leq x < 2$, $\lfloor x \rfloor = 1$. For $2 \leq x < 2.5$, $\lfloor x \rfloor = 2$.
- Substitute the values of floor function and evaluate the integrals
$$ \int_{0}^{1} 0 , dx + \int_{1}^{2} 1 , dx + \int_{2}^{2.5} 2 , dx = 0 + [x]{1}^{2} + [2x]{2}^{2.5} = (2-1) + (2(2.5) - 2(2)) = 1 + (5-4) = 1 + 1 = 2 $$
$ \int_{0}^{1.5} \lfloor x \rfloor dx = 0.5 $
$ \int_{0}^{2.5} \lfloor x \rfloor dx = 2 $
More Information
The floor function, denoted by $\lfloor x \rfloor$, returns the greatest integer less than or equal to $x$. Definite integrals involving the floor function are often evaluated by splitting the integral into intervals where the floor function has a constant value.
Tips
A common mistake is forgetting to split the integral at integer points. Another one is incorrectly evaluating the floor function within each interval. Also, making arithmetic errors, especially when computing the final values.
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