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If $f\left( x \right)\ge g\left( x \right)$ in [a, c] and $f\left( x \right)\le g\left( x \right)$ in [c ,b], where a < c < b, then the area of the regions bounded by the curves in [a, b] is
A. $A=\int_{a}^{c}{\left[ g\left( x \right)-f\left( x \right) \right]}dx+\int_{c}^{b}{\left[ g\left( x \right)+f\left( x \right) \right]}dx$
B. $A=\int_{a}^{c}{\left[ g\left( x \right)-f\left( x \right) \right]}dx+\int_{c}^{b}{\left[ f\left( x \right)-g\left( x \right) \right]}dx$
C. $A=\int_{a}^{c}{\left[ g\left( x \right)+f\left( x \right) \right]}dx+\int_{c}^{b}{\left[ f\left( x \right)+g\left( x \right) \right]}dx$
D. $A=\int_{a}^{c}{\left[ f\left( x \right)-g\left( x \right) \right]}dx+\int_{c}^{b}{\left[ g\left( x \right)-f\left( x \right) \right]}dx$

seo-qna
Last updated date: 22nd Mar 2024
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MVSAT 2024
Answer
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Hint: f(x) will give higher value than g(x) in the domain $x\in \left( a.c \right)$ and vice-versa for domain $x\in \left( c,b \right)$ . Draw functions f(x) and g(x) with the given conditions in given domains. Area under any function h(x) from x = m to ‘n’ can be given as
$\int_{a}^{b}{f\left( x \right)}dx$

Complete step-by-step answer:
we need to determine the area formed by the region of curves f(x) and g(x) with relation as
$\begin{align}
  & f\left( x \right)\ge g\left( x \right)\Rightarrow \left[ a,c \right] \\
 & f\left( x \right)\ge g\left( x \right)\Rightarrow x\in \left[ a,c \right]........................\left( i \right) \\
 & f\left( x \right)\le g\left( x \right)\Rightarrow \left[ c,b \right] \\
 & f\left( x \right)\le g\left( x \right)\Rightarrow x\in \left[ c,b \right].....................\left( ii \right) \\
\end{align}$
With the condition a < c < b. Now, we can observe from the relation (i) that the value of f(x) will be higher than the value of g(x) if ‘x’ will lie in domain [a, c]. In other words a graph of y = f(x) will lie higher than the graph of y = g(x) in the domain $x\in \left[ a,c \right]$. And similarly, the value of g(x) is higher than f(x) for the domain $x\in \left[ c,b \right]$. It means the graph of g(x) will lie upper than the graph of f(x) for $x\in \left[ c,b \right]$ . It means we can draw graphs of function f(x) and g(x) with the relation above as
seo images

Now, as we know area of any function h(x) with x-axis from x = m to x = n can be calculated with the help of integration as
$\text{Area}=\int_{m}^{n}{h\left( x \right)dx....................\left( iii \right)}$
Hence, we can observe the diagram and get that area of bounded regions by f(x) and g(x) are OABO and ADCA. And we can get the area of shaded region OABO can be calculated by the difference of the area of region PBAQP and region POAQ, and similarly, area of ADCA can be calculated by the difference of the area of QACR and QADR. SO, we can get area of shaded region as
Area of shaded region = (area of PBAQP – area of POAQ) + (area of QACRQ – area of QADR)………(iv)
Now, with the help of equation (iii), we can get the area of all the terms involved in the right hand side of the equation (iv). So, we get
\[\begin{align}
  & \text{Area of shaded region =}\int_{a}^{c}{f\left( x \right)}dx-\int_{a}^{c}{g\left( x \right)}dx+\int_{c}^{b}{g\left( x \right)}dx-\int_{c}^{b}{f\left( x \right)}dx \\
 & =\int_{a}^{c}{\left[ f\left( x \right)-g\left[ x \right] \right]dx-\int_{c}^{b}{\left[ g\left( x \right)-f\left( x \right) \right]}dx} \\
\end{align}\]

So, the correct answer is “Option D”.

Note: Understanding the expression
$f\left( x \right)\ge g\left( x \right),f\left( x \right)\ge g\left( x \right)\Rightarrow x\in \left[ a,c \right],x\in \left[ c,b \right]$
And drawing the graph w.r.t are the key points of the question. There should not be an intersection point ‘A’ of both the functions f(x) and g(x). It can be drawn in the following manner as well.
seo images

Where ‘O’ (white circle) is for excluding the point ‘.’ Is used for including the point.