Question

Let $f\left(x\right)=x^3-3x^2+3x+1$ and g be the inverse of f. Then area bounded by y = g(x) and the x-axis from x = 1 to x = 2 is

• A. $\frac{1}{2}sq.unit$
• B. $\frac{1}{4}sq.unit$
• C. $\frac{3}{8}sq.unit$
• D. $\frac{7}{16}sq.unit$

Answer 1

The correct option is B $\frac{1}{4}sq.unit$

$f\left(x\right)=(x-1{)}^3+2\Rightarrow f^{\prime }(x)=3(x-1{)}^2$
x = 1 is a point of inflexion.
So f(x) is strictly increasing in (1, 2)
f and g are symmetric about y = x, hence area bounded by y = g(x) and x-axis from x = 1 to x = 2 is same as area bounded by y = f(x) and y-axis from y = 1 to y = 2.



Required area $=2\times 1-{\int }_0^1(x^3-3x^2+3x+1)dx$

$=2-(\frac{1}{4}-\frac{3}{3}+\frac{3}{2}+1)=\frac{1}{4}sq.unit$