Question

The value of $\left[\int_{0}^{2} [x^2 - x + 1] dx\right]$ where [.] denotes greatest integer function is

Answer 1

1

Answer 2

To evaluate the integral $\left[\int_{0}^{2} [x^2 - x + 1] dx\right]$, we first need to evaluate the definite integral $I = \int_{0}^{2} [x^2 - x + 1] dx$.

Let $f(x) = x^2 - x + 1$. This is a quadratic function. The minimum value of $f(x)$ occurs at $x = \frac{1}{2}$, and the minimum value is $f\left(\frac{1}{2}\right) = \frac{3}{4}$.

Now, let's find the values of $f(x)$ at the endpoints of the interval $[0, 2]$: $f(0) = 1$ and $f(2) = 3$.

The range of $f(x)$ for $x \in [0, 2]$ is $[3/4, 3]$. We need to determine the intervals for $x$ where $[f(x)]$ takes different integer values.

1. For $x \in [0, 1]$: The function $f(x)$ decreases from $f(0)=1$ to $f(1/2)=3/4$, and then increases to $f(1)=1$. So, for $x \in [0, 1]$, the range of $f(x)$ is $[3/4, 1]$. For $x \in (0, 1)$, $3/4 \le f(x) < 1$, which means $[f(x)] = 0$. Therefore, $\int_{0}^{1} [x^2 - x + 1] dx = \int_{0}^{1} 0 dx = 0$.

2. For $x \in [1, 2]$: The function $f(x)$ increases monotonically from $f(1)=1$ to $f(2)=3$. We need to find the point where $f(x)$ crosses the integer value $2$. Set $f(x) = 2$: $x^2 - x + 1 = 2 \implies x^2 - x - 1 = 0$. Using the quadratic formula, $x = \frac{1 \pm \sqrt{5}}{2}$. Since $x \in [1, 2]$, we take the positive root: $x_0 = \frac{1 + \sqrt{5}}{2}$.

   Now we split the integral from $1$ to $2$ at $x_0$:

   • For $x \in [1, x_0)$: $1 \le f(x) < 2$, so $[f(x)] = 1$. $\int_{1}^{x_0} [x^2 - x + 1] dx = \int_{1}^{x_0} 1 dx = [x]_{1}^{x_0} = x_0 - 1 = \frac{1 + \sqrt{5}}{2} - 1 = \frac{\sqrt{5} - 1}{2}$.

   • For $x \in [x_0, 2]$: $2 \le f(x) \le 3$. For $x \in [x_0, 2)$, $2 \le f(x) < 3$, so $[f(x)] = 2$. $\int_{x_0}^{2} [x^2 - x + 1] dx = \int_{x_0}^{2} 2 dx = [2x]_{x_0}^{2} = 2(2) - 2x_0 = 4 - 2\left(\frac{1 + \sqrt{5}}{2}\right) = 3 - \sqrt{5}$.

Now, sum up the contributions from all intervals: $I = \int_{0}^{1} [x^2 - x + 1] dx + \int_{1}^{x_0} [x^2 - x + 1] dx + \int_{x_0}^{2} [x^2 - x + 1] dx$ $I = 0 + \left(\frac{\sqrt{5} - 1}{2}\right) + (3 - \sqrt{5}) = \frac{5 - \sqrt{5}}{2}$.

Finally, we need to find the greatest integer of $I$: $\left[\frac{5 - \sqrt{5}}{2}\right]$. Since $\sqrt{5} \approx 2.236$, $\frac{5 - \sqrt{5}}{2} \approx \frac{2.764}{2} = 1.382$. Therefore, $\left[\frac{5 - \sqrt{5}}{2}\right] = [1.382] = 1$.