Find the area of the region between $y = ( x - 1 ) ^ { 2 }$ and $y = - ( x - 1 ) ^ { 2 }$ for x in $[ 0,1 ]$ .
A) $- \frac { 5 } { 3 }$
B) $\frac { 14 } { 3 }$
C) $\frac { 2 } { 3 }$
D) $\frac { 11 } { 3 }$
E) $- \frac { 2 } { 3 }$

Question

Find the area of the region between $y = ( x - 1 ) ^ { 2 }$ and $y = - ( x - 1 ) ^ { 2 }$ for x in $[ 0,1 ]$ .  
A)  $- \frac { 5 } { 3 }$
B) $\frac { 14 } { 3 }$
C) $\frac { 2 } { 3 }$
D) $\frac { 11 } { 3 }$
E) $- \frac { 2 } { 3 }$

Quizplus · Accepted Answer

The area between the curves $y = (x - 1)^2$ and $y = -(x - 1)^2$ from $x = 0$ to $x = 1$ can be found by integrating the difference of the two functions over the interval $[0, 1]$. The integral is:$$\int_{0}^{1} \left( (x - 1)^2 - (-(x - 1)^2) \right) dx = \int_{0}^{1} 2(x - 1)^2 dx$$Solving the integral:$$= 2 \int_{0}^{1} (x^2 - 2x + 1) dx = 2 \left[ \frac{x^3}{3} - x^2 + x \right]_{0}^{1}$$$$= 2 \left[ \left( \frac{1}{3} - 1 + 1 \right) - \left( 0 - 0 + 0 \right) \right] = 2 \left( \frac{1}{3} \right) = \frac{2}{3}$$Therefore, the area of the region between the given curves over the specified interval is $\frac{2}{3}$, which corresponds to choice C.

Find the Area of the Region Between y=(x−1)2y = ( x - 1 ) ^ { 2 }y=(x−1)2

Find the Area of the Region Between $y = ( x - 1 ) ^ { 2 }$