Sketch the region enclosed by $y = 1 + \sqrt { x }$ and $y = \frac { 4 + x } { 4 }$. Find the area of the region.
Select the correct answer.
A) $\frac { 32 } { 3 }$
B) $\frac { 4 \sqrt { 2 } } { 3 } - 4$
C) $256 \sqrt { 3 }$
D) $\frac { 32 } { 3 } \sqrt { 2 }$
E) $4 \sqrt { 2 }$

Question

Quizplus · Accepted Answer

The correct answer is A.The region is bounded by the curves $y = 1 + \sqrt { x }$ and $y = \frac { 4 + x } { 4 }$. To find the area of the region, we need to find the points of intersection of the two curves. Setting the two equations equal to each other, we get:$$1 + \sqrt { x } = \frac { 4 + x } { 4 }$$$$4 + 4\sqrt { x } = 4 + x$$$$4\sqrt { x } = x$$$$16x = x^2$$$$x^2 - 16x = 0$$$$x(x - 16) = 0$$$$x = 0, 16$$So the two curves intersect at the points $ (0, 1) $ and $ (16, 5) $.The area of the region is given by the integral of the difference of the two curves from $x = 0$ to $x = 16$:$$Area = \int_0^{16} \left ( \frac { 4 + x } { 4 } - (1 + \sqrt { x }) \right ) dx$$$$= \int_0^{16} \left ( \frac { x } { 4 } - \sqrt { x } - \frac { 1 } { 4 } \right ) dx$$$$= \left [ \frac { x^2 } { 8 } - \frac { 2x^{3/2} } { 3 } - \frac { x } { 4 } \right ]_0^{16}$$$$= \frac { 128 } { 8 } - \frac { 2(16)^{3/2} } { 3 } - \frac { 16 } { 4 }$$$$= 16 - \frac { 256 } { 3 } - 4$$$$= \frac { 32 } { 3 }$$Therefore, the area of the region is $\frac { 32 } { 3 }$.

Sketch the Region Enclosed By y=1+xy = 1 + \sqrt { x }y=1+x​

Sketch the Region Enclosed By $y = 1 + \sqrt { x }$