Question
Mathematics Question on applications of integrals
Let the straight line x=b divide the area enclosed by y=(1−x)2 , y=0 and x=0 into two parts R1(0≤x≤b) and R2(b≤x≤1) such that R1−R2=41. Then, b equals
A
43
B
21
C
31
D
41
Answer
21
Explanation
Solution
Given, curve y=(1−x)2 is a parabola, which open upward
∴R1=∫0bydx
=∫0b(1−x)2dx=[−3(1−x)3]0b
=3−1[(1−b)3−1] and R2=∫1by′dx
=∫1b(1−x)2dx=[−3(1−x)3]1b
=−31[−3(1−b)3−0]
=+31[(1−b)3]
But it is given, R1−R2=41
∴−31[(1−b)3−1]−31[(1−b)3]=41
⇒−31[2(1−b)3−1]=41
⇒2(1−b)3=−43+1
⇒(1−b)3=1/8
⇒1−b=1/2
⇒b=1−21
=21