$y(x) = \begin{cases} 2 & \text{if } 0 \le x < \frac{1}{3} \\ 3 & \text{if } \frac{1}{3} \le x < \frac{3}{4} \\ 1 & \text{if } \frac{3}{4} \le x \le 1 \end{cases}$
Which one of the following is the area under the curve for the interval $[0, 1]$ on the x-axis?
The area under the curve of a function defined over an interval is calculated by summing the areas of the sections defined by the function's definition. For a piecewise constant function, this area is the sum of the areas of rectangles, where each rectangle's area is its width multiplied by its height.
The function $y(x)$ is defined over the interval $[0, 1]$ in three parts:
The total area under the curve is the sum of the areas from the three intervals:
$ \text{Total Area} = A_1 + A_2 + A_3 $
$ \text{Total Area} = \frac{2}{3} + \frac{5}{4} + \frac{1}{4} $
First, add the fractions with the same denominator:
$ \frac{5}{4} + \frac{1}{4} = \frac{6}{4} = \frac{3}{2} $
Now, add this result to the area from the first interval:
$ \text{Total Area} = \frac{2}{3} + \frac{3}{2} $
Find a common denominator, which is 6:
$ \text{Total Area} = \frac{2 \times 2}{3 \times 2} + \frac{3 \times 3}{2 \times 3} = \frac{4}{6} + \frac{9}{6} $
$ \text{Total Area} = \frac{4 + 9}{6} = \frac{13}{6} $
The area under the curve for the interval $[0, 1]$ is $\frac{13}{6}$.
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