Area of bounded Regions

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Question

$Consider two curves C_{1} : y^{2} = 4 [\sqrt{y}] x and C_{1} : x^{2} = 4 [\sqrt{x}] y, where [.] denotes G.I.F$ then the area of region enclosed by these two curves within the square formed by the lines $x = 1, y = 1, x = 4 & y = 4 is$

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Solution

$for y \in [1, 4), [\sqrt{y}] = 1 \Rightarrow y^{2} = 4 x$
$for x \in [1, 4), [\sqrt{x}] = 1 \Rightarrow x^{2} = 4 y$
[SolutionStep2]:
$A = \int_{1}^{2} (2 \sqrt{x} - 1) d x + \int_{2}^{4} (2 \sqrt{x} - \frac{x^{2}}{4}) d x$
$= \int_{1}^{2} (2 \sqrt{x} - 1) d x + \int_{2}^{4} (2 \sqrt{x} - \frac{x^{2}}{4}) d x$
$= {(2. \frac{2}{3} x^{\frac{3}{2}} - x)}_{1}^{2} + {(2. \frac{2}{3} x^{\frac{3}{2}} - \frac{x^{3}}{12})}_{2}^{4}$
$= (\frac{4}{3} .2 \sqrt{2} - 2) - (\frac{4}{3} - 1) + (\frac{4}{3} .8 - \frac{64}{12}) - (\frac{4}{3} .2 \sqrt{2} - \frac{8}{12})$
$= \frac{8}{3} \sqrt{2} - 2 - \frac{1}{3} - \frac{8}{3} \sqrt{2} + \frac{8}{12}$
$= \frac{16}{3} + \frac{2}{3} - \frac{7}{3}$
$= \frac{18 - 7}{3} = \frac{11}{3} = 3.67 sq.units$

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Area of bounded Regions

Unlock the full solution & master the concept.

Consider two curves C1:y2=4[y]x and C1:x2=4[x]y, where [.] denotes G.I.F then the area of region enclosed by these two curves within the square formed by the lines x=1,y=1,x=4&y=4 is

for y∈[1,4),[y]=1⇒y2=4x for x∈[1,4),[x]=1⇒x2=4y[SolutionStep2]:A=∫122x−1dx+∫242x−x24dx=∫122x−1dx+∫242x−x24dx=2.23x32−x12+2.23x32−x31224=43.22−2−43−1+43.8−6412−43.22−812=832−2−13−832+812=163+23−73=18-73=113=3.67 sq.units

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$Consider two curves C_{1} : y^{2} = 4 [\sqrt{y}] x and C_{1} : x^{2} = 4 [\sqrt{x}] y, where [.] denotes G.I.F$ then the area of region enclosed by these two curves within the square formed by the lines $x = 1, y = 1, x = 4 & y = 4 is$