Answer :

Roughly plot the curve y = x^{3} and the lines y = x + 6 and x = 0

x = 0 means Y-axis

We have to find the area between the curve y = x^{3} and the line y = x + 6 and Y-axis as shown

Solve y = x + 6 and y = x^{3} to find the intersection point

Put y = x^{3} in y = x + 6

⇒ x^{3} = x + 6

⇒ x^{3} – 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)(x^{2} + 2x + 3) = 0

Observe that x^{2} + 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 = x^{3} 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 = x^{3} on Y-axis – area by y = x + 6 on Y-axis …(i)

Let us find the area under y = x^{3}

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 unit^{2}

Hence area bounded by given curves is 10 unit^{2}

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