Question

Let 3 f x x 3x 2 and g x be its inverse. If the area bounded by g x , x axis and the ordinates x = 2 and x = 6 is p q (where p& q are coprime), find value of p q 1

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Given:

$f(x)=x^{3}+3 x+2$

x = 2 and x = 6

$\text { The complication is the change of sign of } f \text { at } x=g(0) \text {. }$

$\text { We could express } A \text { as }$

$A=A_{1}-\left(-\int_{g(2)}^{g(0)} f(x) d x\right)+A_{2}+\int_{g(0)}^{0} f(x) d x+A_{3}-\int_{0}^{g(6)} f(x) d x$4

$=A_{1}+A_{2}+A_{3}+\int_{g(2)}^{0} f(x) d x-\int_{0}^{g(6)} f(x) d x$

$\text { with } A_{1}=(g(0)-g(2))(-(2)), A_{2}=(0-g(0))(-(2)) \text { and } A_{3}=(g(6)-0) 6 \text {. Their }$

$\text { sum reduces to }$

$A_{1}+A_{2}+A_{3}=-2 g(2)+6 g(6)$

$\text { So we need to know only two values of } g \text { and can avoid the complicated } g(0) \text { : }$

$\begin{array}{l}g(-2):2=f(x)=x^{3}+3 x+2 \Longleftrightarrow x=1 \\g(6): 6=f(x)=x^{3}+3 x+2 \Longleftrightarrow 4=x^{3}+3 x \Longleftrightarrow x=1\end{array}$

$\begin{aligned}A &=8+\int_{1}^{0} f(x) d x-\int_{0}^{1} f(x) d x \\&=8+\left[\frac{1}{4} x^{4}+\frac{3}{2} x^{2}+2 x\right]_{1}^{0}-\left[\frac{1}{4} x^{4}+\frac{3}{2} x^{2}+2 x\right]_{0}^{1}\end{aligned}$

$=8+(7 / 4+2)-(7 / 4+2)$

=8