Question

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle. y = 8x, y = 4x^2.

Answer 1

The area of the region enclosed by the curve y=8x, y=$4x^{2}$ is 64/6 square units.

Given that the curves are y=8x, y=$4x^{2}$.

We are required to find the area of the region enclosed by both the curves.

We have to first find the intersection point of both the curves.

y=8x---------1

y=$4x^{2}$---------2

From 1 & 2

8x=4$x^{2}$

$4x^{2} -8x=0$

4x(x-2)=0

4x=0, x-2=0

x=0,x=2

Use the value of x in 1

y=8*2

=16

Point will be (2,16).

Area will be the area from (0,0) and (2,16). We have to integrate with respect to x and take range from 0 to 2.

Area=$\int\limits^2_0 {8x-4x^{2} } \, dx$

=$[8x^{2} /2-4x^{2} /3]_{0} ^{2}$

=$8*(2)^{2}/2-4(2)^{2} /3-0$

=(8*4)/2-(4*4)/3

=32/2-16/3

=(96-32)/6

=64/6 square units

Hence the area of the region enclosed by the curve y=8x, y=$4x^{2}$ is 64/6 square units.

Learn more about area at brainly.com/question/25965491

#SPJ4