Find the area of the region bounded by: `2y=5sqrtx` , `y=4`   and   `2y+2x=7`

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

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Let,

EQ: `2y=5sqrtx`        EQ2: `y=4`              EQ3: `2y+2x=7`

Then, solve the intersection points between two equations. Use substitution method of system of equations.

> To solve the intersection point betwee EQ1 and EQ2, substitute y=4 to EQ1.

`2y=5sqrtx`                  `8=5sqrtx`                     `64=25x`           

`2*4=5sqrtx`               `8^2=(5sqrtx)^2 `                  `64/25=x`

Since EQ1 is a radical equation, substitute x=64/25 and y=5 to check.

` ``2y=5sqrtx`         `2*4=5sqrt(64/25)`          `8 = 5*8/5`            `8=8`  (True)

So EQ1 and EQ2 intersects at (64/25,4).

> For EQ1 and EQ3, subsitute EQ1 to EQ3.

` ``2y+2x=7`                       `4x^2 -53x + 49 = 0 `       

`5sqrtx+2x=7`                      `(x-1)(4x-49)=0`

`5sqrtx=7-2x`                      `x-1=0` ,    `x=1`

`(5sqrtx)^2=(7-2x)^2`                         and

`25x=49-28x+4x^2`        `4x-49=0` ,   `x=49/4`                

` `

` ` Substitute values of x to EQ3.

` ``x=1` ,        `2y+2*1=7`        `2y+2= 7`        `2y=5`           `y=5/2`        ` `         ` `      ` `          ` `

`x=49/4` ,    `2y+2*49/4=7`       `2y+49/2=7`       `2y=-35/2`       `y=-35/4`    ` `        ` `     ` `     ` `

Since EQ1 is a radical equation, substitute values of x and y to EQ1.

`(1,5/2)` ,                   `2*5/2=5sqrt1`           `5=5*1`           `5=5`  (True)         

` ` `(49/4.-35/4)` ,   `2*(-35/4)=5sqrt(49/4)`     `-35/2=5*7/2`     `-35/2=35/2` ( False)

Hence, EQ1 and EQ2 intersects at (1,5/2).

>   For EQ2 and EQ3, substitute y=4 to EQ3.

`2y+2x=7`              `8+2x=7`           `x = -1/2`                   

`2*4+2x=7`               `2x=-1`  

Thus, EQ1 and EQ2 intersects at (-1/2,4).    

So the region bounded by the three equations is:

(Note: Blue- Graph of EQ1, Red-Graph of EQ2, and Green-Graph of EQ3.) 

Then, let's use the formula for area between curves.

`A =int_a^b (y_U-y_L)dx`

And base on the graph above, there is one upper function (`y_U=4` ) and two lower functions (`y_(L1)=(7-2x)/2` and `y_(L2)=(5sqrtx)/2` ) .Also, the limits of the integral are the x-coordinates of the three intersection point.

So we have:

`A = int_a^b (y_U-y_(L1))dx + int_b^c (y_u-y_(L2))dx`

`A=int_(-1/2)^1(4-(7-2x)/2)dx + int_1^(64/25)(4-(5sqrtx)/2)dx`

`A= 1/2int_(-1/2)^1 (1+2x)dx + 1/2int_1^(64/25)(8-5sqrtx) dx`

`A= 1/2(x+x^2)` `|_(-1/2)^1`    `+`     `1/2(8x - (10xsqrtx)/3)` `|_1^(64/25)`

`A=9/8 + 27/5`

`A= 441/200 = 2.205`

Hence, the area bounded by the three equations is 2.205 square units.

Sources:

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