GATE CSE 2022 | GA | Question: 2

Question

A function $y(x)$ is defined in the interval $[0, 1]$ on the $x – $ axis as

$$y(x) = \left\{\begin{matrix} 2& \text{if} & 0 \leq x < \frac{1}{3} \\ 3& \text{if}& \frac{1}{3} \leq x < \frac{3}{4} & \\ 1 & \text{if} & \frac{3}{4} \leq x \leq 1& \end{matrix}\right.$$

Which one of the following is the area under the curve for the interval $[0, 1]$ on the $x – $ axis?

• A. $\frac{5}{6}$
• B. $\frac{6}{5}$
• C. $\frac{13}{6}$
• D. $\frac{6}{13}$

Answer 1

Answer: Option C.

The Area of the function $y(x)$ is composed of an area of three rectangles, as shown in the above picture.

$\text{Total area} = \left[2\ast\left(\frac{1}{3} – 0 \right)\right] + \left[3\ast \left(\frac{3}{4}- \frac{1}{3} \right)\right] + \left[1\ast \left(1- \frac{3}{4} \right)\right]$

$\qquad  \qquad \quad =  \frac{2}{3} + \frac{15}{12} + \frac{1}{4} = \frac{26}{12} = \frac{13}{6}\;\text{unit}^{2}.$

Comments

However, area calculated is correct but the graph is not drawn correctly.
One has to clearly specify the value of y for different values of x  i.e. different interval break points such as 1/3 or  3/4 .

For the correct graph, see the answer given by @vermavijay1986

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However, area calculated is correct but the graph is not drawn correctly.

One has to clearly specify the value of y for different values of x  i.e. different interval break points such as 1/3 or  3/4 .

I agree. These are some technicalities which I assume the reader can understand from the question.

Another technicality is that the (correct)answer will be :

 The Area of the function $y(x)$ is composed of an area of three rectangles, as shown in the above picture.

$\text{Total area} = \left[2\ast\left(\frac{1}{3} –h- 0 \right)\right] + \left[3\ast \left(\frac{3}{4}-h- \frac{1}{3} \right)\right] + \left[1\ast \left(1- \frac{3}{4} \right)\right]$ where $h \rightarrow 0$

But for this particular equation, the limit doesn’t affect the calculation.

$\qquad  \qquad \quad =  \frac{2}{3} + \frac{15}{12} + \frac{1}{4} = \frac{26}{12} = \frac{13}{6}\;\text{unit}^{2}.$

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The graph assumes that the transition between <1/3 to 1/3 and <3/4 to 3/4 is negligible and the shape obtained is very close to a rectangle.

Answer 2

Here, option C is correct that can be easily verified by drawing the graph:

Comments

No excluded point in graph according to the quuestion

Answer 3

Area under curve in [0,1] = Area of Rectangle I + Area of Rectangle II + Area of Rectangle III

= (1/3) x 2 + (5/12) x 3 + (1/4) x 1

= 13/6

Answer: C