Definite Integration Question 439
Question: The area enclosed by the curves $ y=sinx+cosx $ and $ y=| \text{cos }x-\text{sin }x | $ over the interval [0, $ \pi /2 $ ] is
Options:
A) 2/3 sq. units
B) 8/3 sq. units
C) 11/3 sq. units
D) 13/6 sq. units
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ A_1=\int\limits_0^{1}{( 1+\sqrt{x}-\frac{x}{4} )dx} $
$ ={{[ x+\frac{2{x^{3/2}}}{3}-\frac{x^{2}}{8} ]}_0}^{1}=1+\frac{2}{3}-\frac{1}{8}=\frac{37}{24}. $
$ A_2=\int\limits_1^{4}{( \frac{2}{\sqrt{x}}-\frac{x}{4} )dx} $
$ ={{[ 4\sqrt{x}=\frac{x^{2}}{8} ]}_1}^{4} $
$ =[ 8-2-4+\frac{1}{8} ] $
$ =\frac{17}{8} $ Or $ A=A_1+A_2=\frac{88}{24}=\frac{11}{3} $ sq. units
BETA
