`y_1 = x^2 , y_2 = x^3` Set up the definite integral that gives the area of the region

Given the curve equations ,they are

`y1=x^2-` -----(1)

`y2=x^3` -----(2)

to get the boundaries or the intersecting points of the curves we have to equate the curves .

y_1=y_2

=> `x^2 = x^3`

=> `x^2-x^3 =0`

=> `x^2(1-x)=0`

=> `x=0 or x=1`

When ` 0<=x<=1 , x^2 > x^3` and so

the Area = `int (x^2 -x^3) dx ` [from 0 to 1]

= `int _0 ^1 (x^2 -x^3) dx `

= `[x^3/3 -x^4/4]_0 ^1`

= `[1/3 -1/4]-[0-0] = (4-3)/12 = 1/12`

so area of the region enclosed by the curves is = `1/12`

Comments

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