Find the area bounded by the curve `y=sqrt(x)` ; `y = (5-x)/4` ; `y=(3x-8)/2`

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lemjay | High School Teacher | (Level 3) Senior Educator

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Let,

EQ.1:  `y=sqrtx`           EQ.2: `y= (5-x)/4`             EQ.3: `y = (3x-8)/2`

Determine the intersection points of the three equations using substitution method.

Susbstitute `y=sqrtx` to EQ.2 .

 `sqrtx= (5-x)/4`

 `4sqrtx = 5-x`

Square both sides.

 ` 16x = 25-10x+x^2`

    `0 = 25 - 26x + x^2`

    `0 = (x-25)(x-1)`

So values of x are:  x = 1    and     x = 25

Substitute the values of x to EQ.1 and EQ.2.

`x=1` ,    `y = sqrt1 = 1 `      and     `y =(5-1)/4 = 1`

`x=25` ,   `y = sqrt25 = 5`    and     `y = (5-25)/4 = -5`

Note that the value of x that satisfy both equation is x=1. Hence, EQ.1 and EQ.2 intersect at (1,1).

Next, substitute `y=sqrtx` to EQ.3.

   `sqrtx = (3x-8)/2`

 `2sqrtx = 3x-8`

Square both sides.

  ` 4x = 9x^2-48x+64`

    ` 0= 9x^2-52x+64`

Using the quadratic formula, the values of x are: x=4 and x=16/9. Then, substitute the values of x to EQ.1 and EQ.3.

`x=4` ,     `y =sqrt4 = 2 `         and        `y=[(3*4)-8]/2 =2`

`x=16/9` ,   `y = sqrt(16/9) =4/3`       and        `y = [(3*16/9)-8]/2 = -4/3`

Base on this, EQ.1 and EQ.3 intersect at (4,2).

Next, substitute y = (5-x)/4  to EQ.3.

     `(5-x)/4 = (3x-8)/2`

`2(5-x) = 4(3x-8)`

 `10-2x = 12x - 32`

      `42 = 14x`

        `3 = x `  

Substitute the value of x to EQ.2 and EQ.3 .

`x = 3` ,    `y =(5-3)/4 = 1/2 `       and     `y = (3(3)-8)/2 = 1/2`

Hence, EQ.2 and EQ.3 intersect at (3,1/2).

Then, plot the three equations.

(Note:Graph of EQ.1- Red; EQ.2-Green ; and EQ.3 - Blue)

To find the area bounded by the three equations, use the formula:

`A = int_a^b( y_U-y_L) dx `  

Base on the graph above, the upper function is `y_U=sqrtx` . And there are two lower functions which are `y_L=(5-x)/4` and y`_L=(3x-8)/2` . Also, the limits of the integral are the x-coordinates of the intersection points. So, the area is:

`A = int_1^3(sqrtx-(5-x)/4)dx + int_3^4(sqrtx-(3x-8)/2)dx`

`A = int_1^3 (sqrtx -5/4 +x/4 )dx + int_3^4(sqrtx-(3x)/2 +4)dx`

`A = (2/3xsqrtx-(5x)/4+x^2/8)|_1^3 + (2/3xsqrtx - (3x^2)/2 +4x)|_3^4`

`A = (12sqrt3-13)/6 + (49-24sqrt3)/12 = 23/12 = 1.9`

Hence, the area bounded by` y=sqrtx` , `y=(5-x)/4` and `y = (3x-8)/2` is 1.9 square units.

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