Let f: R - R be a function defined by f(x) = { [x], x ≤ 2; 0, x 2 where [x] is the greatest integ

Q: Why do we split the integral at $\sqrt{2}$?

The function $f(x)$ is defined piecewise, changing its behavior at $x=2$. When we have $f(x^2)$, the transition point occurs when $x^2=2$, which means $x=\sqrt{2}$. We must split the integral to correctly apply the definition of $f(x)$ in each sub-interval.

Q: What happens to $f(x+1)$ when $x$ is in $[1, \sqrt{2}]$?

In the interval $[1, \sqrt{2}]$, $x$ is greater than $1$, so $x+1$ is greater than $2$. According to the definition $f(x) = 0$ for $x > 2$, $f(x+1)$ becomes $0$. This simplifies the denominator significantly.

Q: Is $f(x^2)$ always $[x^2]$?

No, it is only $[x^2]$ when $x^2 \le 2$. If $x^2 > 2$, the function definition dictates that $f(x^2) = 0$. Always check the range of the argument before applying the greatest integer function.

Question

Let $f: R 	o R$ be a function defined by $f(x) = \begin{cases} [x], & x \le 2 \ 0, & x > 2 \end{cases}$ where $[x]$ is the greatest integer less than or equal to $x$, if $I = \int_{-1}^{2} \frac{x f(x^2)}{2 + f(x+1)} dx$, then the value of $(4I - 1)$ is

Sigma Percentile · Accepted Answer

### Analyzing the Function and Thresholds

The given function is defined as:
$$f(x) = 
\begin{cases} 
[x] & 	ext{for } x \le 2 \ 
0 & 	ext{for } x > 2 
\end{cases}$$

We are tasked with evaluating the integral:
$$I = \int_{-1}^{2} \frac{x f(x^2)}{2 + f(x+1)} dx$$

The behavior of the function changes at the threshold $x=2$. We must identify the critical points where the arguments $x^2$ and $x+1$ cross this threshold:
1. For $x^2 = 2$, we find $x = \sqrt{2}$ (considering the positive domain).
2. For $x+1 = 2$, we find $x = 1$.

Thus, we must partition the integral into intervals based on these critical points: $[-1, 1)$, $[1, \sqrt{2})$, and $[\sqrt{2}, 2]$.

### Evaluating the Intervals

**Interval 1: $x \in [-1, 1)$**
In this range, $x^2$ lies between $0$ and $1$. Since $x^2 \le 2$, we have $f(x^2) = [x^2] = 0$.
Consequently, the numerator $x \cdot f(x^2) = 0$, making the integral over this interval equal to $0$.

**Interval 2: $x \in [1, \sqrt{2})$**
Here, $x^2$ is between $1$ and $2$. Since $x^2 \le 2$, we have $f(x^2) = [x^2] = 1$.
For the denominator, $x+1$ lies between $2$ and $1+\sqrt{2}$. Since $x+1 > 2$, the definition of the function yields $f(x+1) = 0$.
The integrand simplifies to:
$$\frac{x \cdot 1}{2 + 0} = \frac{x}{2}$$

**Interval 3: $x \in [\sqrt{2}, 2]$**
In this range, $x^2$ is between $2$ and $4$. Since $x^2 > 2$, we have $f(x^2) = 0$.
The numerator becomes $x \cdot 0 = 0$, so the integral over this interval is $0$.

### Final Calculation

We are left with the integral over the second interval:
$$I = \int_{1}^{\sqrt{2}} \frac{x}{2} dx$$

Evaluating this using the power rule:
$$I = \left[ \frac{x^2}{4} ight]_{1}^{\sqrt{2}} = \frac{(\sqrt{2})^2}{4} - \frac{1^2}{4} = \frac{2}{4} - \frac{1}{4} = \frac{1}{4}$$

Finally, we compute the requested value:
$$4I - 1 = 4\left(\frac{1}{4}ight) - 1 = 1 - 1 = 0$$

The final result is **$0$**.

Animated Solution for Mathematics - Definite Integration: Let $f : R \to R$ be a function defined by $f (x) = {[x], 0, x \leq 2 x > 2$ where $[x]$ is the greatest integer less than or equal to $x$ , if $I = \int_{- 1}^{2} \frac{x f ( x ^{2} )}{2 + f ( x + 1 )} d x$ , then the value of $(4 I - 1)$ is

Enter Numerical Value:

Visualized Solution

Understanding the function $f (x)$

Analyzing Interval $x \in [- 1, 1)$

Analyzing Interval $x \in [1, 2)$ : Numerator

Analyzing Interval $x \in [1, 2)$ : Denominator

Analyzing Interval $x \in (2, 2]$

Evaluating the Integral $I$

Final Calculation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Function and Thresholds

Evaluating the Intervals

Final Calculation

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Enter Numerical Value:

Visualized Solution

Understanding the function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Analyzing Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈[−1,1)

Analyzing Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈[1,2​): Numerator

Analyzing Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈[1,2​): Denominator

Analyzing Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈(2​,2]

Evaluating the Integral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I

Final Calculation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Function and Thresholds

Evaluating the Intervals

Final Calculation

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Analyzing Interval $x \in [1, 2)$ : Numerator

Analyzing Interval $x \in [1, 2)$ : Denominator

Analyzing Interval $x \in (2, 2]$

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