Let f ( x ) = x 3 + 3 x + 2 and g ( x ) is the inverse of it. The area bounded by g ( x ) , the x -axis and the ordinates at x = − 2 and x = 6 is n m where m , n ∈ N & G . C . D of ( m , n ) = 1 the

Question

Let f ( x ) = x 3 + 3 x + 2 and g ( x ) is the inverse of it. The area bounded by g ( x ) , the x -axis and the ordinates at x = − 2 and x = 6 is n m ​ where m , n ∈ N & G . C . D of ( m , n ) = 1 then m − 2 =

TestHub · Accepted Answer

The required area will be equal to area enclosed by $y = f (x)$ , the y-axis between the abscissa at $y = - 2$ and $y = 6$ Required area

$= \int_{0}^{1} {6 - f (x)} d x + \int_{- 1}^{0} [f (x) - (- 2)] d x$

$\frac{9}{2} = \frac{m}{n} \Rightarrow m - n = 7$

Mathematics - Area Under Curves Question with Solution | TestHub

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