Question

pls help show steps:
find the area of the region
between y=sqrt(x+3). y=(x+3)/2

Answer 1

\[\sqrt{x+3}= \frac{x+3}{2} --> x+3 = \frac{1}{4} (x+3)^2\]

Answer 2

\[\frac{1}{4} ( x^2 +6x +9) -x-3 =0\]
Multiply by 4

\[x^2 +6x+9 -4x-12 =0 --> x^2 +2x -3 =0 \]

Answer 3

(x+3)(x-1) =0 ---> x=-3, 1

Answer 4

So now I can redraw the picture.

Answer 5

Before I drew the line, now I drew in the sqrt graph and shaded area between is what you to find.

Answer 6

so the interval for the integration is -3<=x<=1 and from the picture we can see that the sqrt function is above the line

Answer 7

\[Area= \int\limits_{-3}^{1} \sqrt{x+3} - \frac{x+3}{2} dx \]

Answer 8

so

= int (x+3)^(1/2) - (x+3)/2 dx
= (2/3) (x+3)^(3/2) - (x^2 / 4) - (3x/2) between the limits.

Answer 9

\[= [ \frac{2}{3} (0)^{\frac{3}{2}} - \frac{9}{4} +\frac{9}{2} ] -[ \frac{2}{3} (4)^{\frac{3}{2}} - \frac{1}{4}- \frac{3}{2} ]\]

Put into a calculator.