NIMCET 2018 — Mathematics PYQ

- Area under a curve:
$$\text{The area under the function }y=f(x)\text{ from }x=a\text{ to }x=b\text{ and the x-axis is given by the definite integral }{\int }_a^{b}f(x)dx\text{, for curves which are entirely on the same side of the x-axis in the given range.}$$
$$</span><br><spanstyle="font-family:arial,helvetica,sans-serif;font-size:14pt;">\text{If the curves are on both the sides of the x-axis, then we calculate the areas of both the sides separately and add them.}$$
- Two curves f(x, y) = 0 and g(x, y) = 0 cut/touch at a point (a, b) if f(a, b) = g(a, b) = 0.
- Definite integral:
$$\text{If }\int f(x)dx=g(x)+C,\text{ then }{\int }_a^{b}f(x)dx=[g(x){]}_a^b=g(b)-g(a).$$
$$</span><br><spanstyle="font-family:arial,helvetica,sans-serif;font-size:14pt;">\int x^{n}dx=\frac{x^{n+1}}{n+1}+C.$$

Calculation:

On solving for the common points of the curves $y^2=x$ and $y=|x|$, we find that they intersect at the points (1, 1) and (0, 0), as shown below:

The required area is:

(Area under $y=\sqrt{x}$ from 0 to 1) - (Area under $y=x$ from 0 to 1)
$=\left|{\int }_0^1\sqrt{x}dx\right|-\left|{\int }_0^{1}xdx\right|$
$={\left[\frac{x^{1+1}}{1+1}\right]}_0^1-{\left[\frac{x^{1+1}}{1+1}\right]}_0^1$
$=\frac{2}{3}[1-0]-\frac{1}{2}[1-0]$
$=\frac{1}{6}$