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Question

A function $y(x)$ is defined in the interval $[0, 1]$ on the x-axis as
$y(x) = \begin{cases} 2 & \text{if } 0 \le x < \frac{1}{3} \\ 3 & \text{if } \frac{1}{3} \le x < \frac{3}{4} \\ 1 & \text{if } \frac{3}{4} \le x \le 1 \end{cases}$
Which one of the following is the area under the curve for the interval $[0, 1]$ on the x-axis?

The correct answer is
$\frac{13}{6}$

Calculating Area Under a Piecewise Function

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.

Area Calculation Steps

The function $y(x)$ is defined over the interval $[0, 1]$ in three parts:

  • Interval 1: $0 \le x < \frac{1}{3}$
    • Function value (height): $y(x) = 2$
    • Interval width: $\Delta x_1 = \frac{1}{3} - 0 = \frac{1}{3}$
    • Area 1: $A_1 = \text{height} \times \text{width} = 2 \times \frac{1}{3} = \frac{2}{3}$
  • Interval 2: $\frac{1}{3} \le x < \frac{3}{4}$
    • Function value (height): $y(x) = 3$
    • Interval width: $\Delta x_2 = \frac{3}{4} - \frac{1}{3} = \frac{9}{12} - \frac{4}{12} = \frac{5}{12}$
    • Area 2: $A_2 = \text{height} \times \text{width} = 3 \times \frac{5}{12} = \frac{15}{12} = \frac{5}{4}$
  • Interval 3: $\frac{3}{4} \le x \le 1$
    • Function value (height): $y(x) = 1$
    • Interval width: $\Delta x_3 = 1 - \frac{3}{4} = \frac{1}{4}$
    • Area 3: $A_3 = \text{height} \times \text{width} = 1 \times \frac{1}{4} = \frac{1}{4}$

Total Area

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 $\frac{13}{6}$.

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Important Questions from Area Under Curve

  1. 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 $I$?
  2. 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)

  3. The value of $\frac{4}{\pi} \int_0^{\pi/2} \sin^2 x \text{ dx}$ is _________________ (rounded off to two decimal places).

  4. 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

  5. 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

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