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  • Using integration find the area of the region : <img alt="\left \{ (x, y) : 0 \leq y \leq x ^2 , 0 \leq y \leq x, 0 \leq x \leq 2 \right \}" src="https://entrancecorner.codecogs.com/gif.latex?%5Cleft%20%5C%7B%20%28x%2C%20y%29%20%3A%200%20%5Cleq%20

Using integration find the area of the region : \left \{ (x, y) : 0 \leq y \leq x ^2 , 0 \leq y \leq x, 0 \leq x \leq 2 \right \}

 

 

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\left \{ (x, y) : 0 \leq y \leq x ^2 , 0 \leq y \leq x, 0 \leq x \leq 2 \right \}

Intersect point

y= x ^2,...(i)

y= x

Put value of y in equation (i)

y= y^2

y  = 0 , 1

x = 0,1

Intersect ppiint A(1,1)

\\$Required Area $ = \int _0^{1} x^2 dx + \int _1^{2} x dx \\\\ =[\frac{x^3}{3}] _0^{1}+[\frac{x^2}{2} ] _1^{2} \\\\ =\frac{1}{3}+\frac{3}{2} \\\\ = \frac{2+9}{6} \\\\ = \frac{11}{6} \ square \ units

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Safeer PP

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