Definite Integration Question 153
Question: The area of the figure bounded by $ y^{2}=2x+1 $ and $ x-y=1 $ is
Options:
A) $ \frac{2}{3} $
B) $ \frac{4}{3} $
C) $ \frac{8}{3} $
D) $ \frac{16}{3} $
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Answer:
Correct Answer: D
Solution:
[d] Area of the region is given by $ A=\int\limits _{-1}^{3}{[ (y+1)-( \frac{y^{2}-1}{2} ) ]dy=\frac{16}{3}} $
BETA
