Definite Integration Question 191
Question: The area bounded by $ f(x)=x^{2},0\le x\le 1, $
$ g(x)=-x+2,1\le x\le 2 $ and $ x-axis $ is
Options:
A) $ \frac{3}{2} $
B) $ \frac{4}{3} $
C) $ \frac{8}{3} $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Required area = Area of OAB + Area of ABC Now, Area of $ OAB=\int\limits_0^{1}{f(x)dx+\int\limits_1^{2}{g(x)dx}} $
$ =\int\limits_0^{1}{x^{2}dx+\int\limits_1^{2}{(-x+2)dx=. \frac{x^{3}}{3} |_0^{1}+[ \frac{-x^{2}}{2}+2x ]}_1^{2}} $
$ =\frac{1}{3}+[ ( \frac{-4}{2}+4 )-( \frac{-1}{2}+2 ) ]=\frac{1}{3}+\frac{1}{2}=\frac{5}{6} $ sq. unit
BETA
