Definite Integration Question 428
Question: Let $ f(x)=x^{3}+3x+2 $ and g(x) be the inverse of it. Then the area bounded by g(x), the x-axis, and the ordinate at $ x=-2 $ and $ x=6 $ is
Options:
A) 1/4 sq. units
B) 4/3 sq. units
C) 5/4 sq. units
D) 7/3 sq. units
Show Answer
Answer:
Correct Answer: C
Solution:
[c] The required area will be equal to the area enclosed by $ y=f(x), $ y-axis between the abscissa at $ y=-2 $ and $ y=6 $
Hence, $ A=\int\limits_0^{1}{( 6-f(x) )}dx+\int\limits _{-1}^{0}{( f(x)-(-2) )}dx $
$ =\int\limits_0^{1}{( 4-x^{3}-3x )dx+\int\limits _{-1}^{0}{( x^{3}+3x+4 )}dx=\frac{5}{4}} $ Sq. units.
BETA
