Greatest Integer Function Area Calculation on $[1,4]$
The function is defined as $y = [x]$, where $[x]$ represents the greatest integer less than or equal to $x$. The task is to find the area under this curve for the interval $x \in [1, 4]$.
Area Breakdown by Interval
The interval $[1, 4]$ can be divided into subintervals where the value of $[x]$ remains constant:
- Interval $[1, 2)$: Here, $y = [x] = 1$. The area contribution is calculated as Base $\times$ Height = $(2-1) \times 1 = 1 \times 1 = 1$.
- Interval $[2, 3)$: Here, $y = [x] = 2$. The area contribution is Base $\times$ Height = $(3-2) \times 2 = 1 \times 2 = 2$.
- Interval $[3, 4)$: Here, $y = [x] = 3$. The area contribution is Base $\times$ Height = $(4-3) \times 3 = 1 \times 3 = 3$.
Total Area Computation
The total area under the curve $y = [x]$ from $x=1$ to $x=4$ is the sum of the areas calculated for each subinterval. This can be represented as the definite integral:
$ \text{Total Area} = \int_{1}^{4} [x] \, dx = \int_{1}^{2} 1 \, dx + \int_{2}^{3} 2 \, dx + \int_{3}^{4} 3 \, dx $
Summing the individual areas:
$ \text{Total Area} = 1 + 2 + 3 = 6 $
Therefore, the area under $y = [x]$ for $x \in [1, 4]$ is 6 square units.
Important Questions from Area Under Curve
- Let $I$ be the integral defined as follows: $$I = \int_{0}^{1} \int_{0}^{\sqrt{y}} dx dy + \int_{1}^{2} \int_{\sqrt{y-1}}^{1} dx dy$$ If the order of the integration is changed, then which one of the following is the correct expression for $I$?
Let $\alpha = \iint_S \vec{F} \cdot \hat{n} \, dS$, where $\vec{F} = (2x + 3z)\hat{i} + (xz - y)\hat{j} + (y^2 + 2z)\hat{k}$ and $S$ is the sphere with centre at $(3, -1, 2)$ and radius 9. Here, $\hat{n}$ is the unit normal drawn outward and $\hat{i}, \hat{j}, \hat{k}$ are unit vectors.
Then the value of $\frac{1}{36\pi} \alpha$ is equal to ________. (answer in integer)
The value of $\frac{4}{\pi} \int_0^{\pi/2} \sin^2 x \text{ dx}$ is _________________ (rounded off to two decimal places).
The work done by the force $F = (x + y)\hat{i} - (x^2 + y^2)\hat{j}$, where $\hat{i}$ and $\hat{j}$ are unit vectors in $\vec{OX}$ and $\vec{OY}$ directions, respectively, along the upper half of the circle $x^2 + y^2 = 1$ from $(1,0)$ to $(-1,0)$ in the $xy$-plane is
If the line $y = \alpha x$, $\alpha \geq \sqrt{2}$, divides the area of the region
$R: = \{(x, y) \in \mathbb{R}^2| 0 \leq x \leq \sqrt{y}, 0 \leq y \leq 2\}$
into two equal parts, then the value of $\alpha$ is equal to
