Find the area of the region bounded by the curve y = x3 and y = x + 6 and x = 0.
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