Definite Integration Question 194
Question: What is the area bounded by the curve $ y=4x-x^{2}-3 $ and the x-axis-
Options:
A) 2/3 sq. unit
B) 4/3 sq. unit
C) 5/3 sq. unit
D) 4/5 sq. unit
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Given curve is $ y=4x-x^{2}-3 $ Since, area bounded by x-axis
$ \therefore y=0 $
$ \Rightarrow 4x-x^{2}-3=0\Rightarrow x^{2}-4x+3=0 $
$ \Rightarrow x^{2}-3x-x+3=0\Rightarrow (x-3)(x-1)=0 $
$ \Rightarrow x=1,3 $
$ \therefore $ Required area $ =\int_1^{3}{(4x-x^{2}-3)dx} $
$ . =\frac{4x^{2}}{2}-\frac{x^{3}}{3}-3x |_1^{3}=( \frac{36}{2}-\frac{27}{3}-9 ) $
$ -( \frac{4}{2}-\frac{1}{3}-3 ) $
$ =(18-9-9)-( 2-\frac{10}{3} )=0-( \frac{-4}{3} )=\frac{4}{3} $ sq. unit.
BETA
