#### To determine

**To find:**

The relationship between A and B

#### Answer

B=16A

#### Explanation

**1) Concept:**

Area between curve:

The area A of the region bounded by curves y=fx, y=gx and the line x=c, x=d, where f and g are continuous and f(x)≥g(x) for all x in [c, d], is

A=∫cdfx-gxdx

**2) Calculation:**

Assume P lies on the curve and in first quadrant.

As y=x3 lies in the first and third quadrant, this assumption is valid.

Let P=a, a3.

Use the below information to find the slope of the tangent line to curve y=x3 at the point P=(a, a3)

Here, fx=x3 and fa=a3, so the slope is

m=limx→afx-fax-a=limx→ax3-a3x-a

=limx→a(x-a)(x2+xa+a2)x-a

=limx→a(x2+xa+a2)=3a2

Using point slope form, find the equation of tangent line at a, a3.

y-y1=m(x-x1)

y-a3=3a2(x-a)

y=3a2x-2a3

Now plot the curve and tangent line to find the point Q.

It is the intersection of the curve y= x3 and tangent line y=3a2x-2a3.

To find the point Q:

x3=3a2x-2a3

x3-3a2x+2a3=0

x-a2x+2a=0

So x=a and x= -2a

Our point P corresponds to the first value x=a

So required x=-2a, y=-2a3=-8a3

So point Q is (-2a, -8a3).

Now we need to find the slope of the tangent line to the given curve passing through Q.

Here, fx=x3 and fa=-8a3, so the slope is

m=limx→afx-fax-a=limx→-2ax3+8a3x+2a

=limx→-2a(x+2a)(x2-2xa+4a2)x+2a

=limx→-2a(x2-2xa+4a2)=12a2

Using the point slope form, find the equation of the tangent line at (-2a,-8a3).

y-y1=m(x-x1)

y+8a3=12a2(x+2a)

y=12a2x+16a3

To find the point P2, which is the intersection of this tangent line at Q and the curve y=x3.

x3=12a2x+16a3

x3-12a2x-16a3=0

x+2a2x-4a=0

So, x=4a, y=4a3=64a3

So point P2 is (4a, 64a3).

Since we need to find the relation between area A and B, write the area A in terms of integration.

A=∫cdfx-gxdx

Here, c=-2a to d=a, fx=x3 and gx=3xa2-2a3

A=∫-2aax3-(3xa2-2a3)dx

A=x44-32a2x2+2a3x-2aa

A=a44-32a2a2+2a3a--2a44-32a2-2a2+2a3(-2a)

A=a44-32a4+2a4-16a44+122a4+4a4

A=-15a44+9a42+6a4

A=-15a4+18a4+24a44

A=27a44

Now to find area B.

B=∫cdfx-gxdx

Here, c=-2a to d=4a, fx=12xa2+16a3 and gx=x3

B=∫-2a4a12xa2+16a3-x3dx

B=12a2x22+16a3x-x44-2a4a

B=192a42+64a4-256a44-48a42-32a4-16a44

B=96a4+64a4-64a4-24a4+32a4+4a4

B=108a4

To find the relationship between A and B, divide A by B.

AB=27a44108a4

AB=116

B=16A

So, area of B is 16 times area of A.

**Conclusion:**

B=16A