Answer :

Roughly plot the curve y = x3 and the lines y = x + 6 and x = 0


x = 0 means Y-axis


We have to find the area between the curve y = x3 and the line y = x + 6 and Y-axis as shown


Solve y = x + 6 and y = x3 to find the intersection point


Put y = x3 in y = x + 6


x3 = x + 6


x3 – x – 6 = 0


Mentally checking if 0,1,2 satisfy the cubic we get that one root is 2 hence x – 2 is a factor


Take that factor out



(x – 2)(x2 + 2x + 3) = 0


Observe that x2 + 2x + 3 don’t have real roots


Hence x = 2


Put x = 2 in y = x + 6 we get y = 8


Hence both curves y = x3 and y = x + 2 intersect at (2, 8)



As we have to find area on Y-axis we should integrate x = f(y) that is here we are taking a horizontal strip of length dy


So the area bounded will be


area bounded = area by y = x3 on Y-axis – area by y = x + 6 on Y-axis …(i)



Let us find the area under y = x3


As we need in terms of x = f(y)



Integrate from 0 to 8











Now let us find area under y = x + 6 on Y axis


Observe in figure that we have to find area from 6 to 8 because the line intersects Y-axis at 6 and upto 8 because that is the y-coordinate where the curve and line intersects


x = y – 6


Integrate from 6 to 8







Using (i)


area bounded = 12 – 2 = 10 unit2


Hence area bounded by given curves is 10 unit2


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