   Chapter 4.P, Problem 2P

Chapter
Section
Textbook Problem

# Find the minimum value of the area of the region under the curve y = 4 x − x 3 from x = a to x = a + 1 , for all a > 0.

To determine

To find:

The minimum value of the area under the curve

Explanation

1) Concept:

Power rule of integration:

abf(x)dx=xn+1n+1ab

abfx-gxdx=abfxdx-abgxdx

2) Given:

y=4x-x3

3) Calculation:

Here,

y=4x-x3

From x=a to x=a+1, for all a>0

We shall consider are three cases

i. 0<a1

ii. 1<a<2

iii. a 2

For,0<a1

The value of the area under the curve is

aa+1ydx=aa+14x-x3dx

=aa+14xdx-aa+1x3dx

=4aa+1xdx-aa+1x3dx

Using power rule of integration,

=4x22aa+1-x44aa+1

=4a+12-a22-a+14-a44

=4a+12-a22-a+12-a2a+12+a24

=a+12-a248-a+12+a2

=a2+2a+1-a248-a2+2a+1+a2

=2a+148-2a2-2a-1

=2a+14-2a2-2a+7

=-4a3-4a2+14a-2a2-2a+74 So

aa+1ydx=A=-4a3-6a2+12a+74

A=-4a3-6a2+12a+74

To find the where the area is maximum or minimum, check critical points

That is solve the equation A'=0

Differentiate the area with respect to  a,

A'=14-12a2-12a+12

Consider A'=0

-12a2-12a+12=0

By dividing -12,

a2+a-1=0

a=-1±52

It is given that, a>0

So, a=5-12

Now again the differentiate the area with respect to a,

A''=14-24a-12

So substitute  a=5-12,

A''=14-245-12-12

=14-125-1-12

=14-125

=-35

As the second derivative is negative, this area (at a=5-12) is a local maximum.So the minimum in the interval [0,1] shall occur at a=1  or a=0. At a=0 area is A(0)=7/4=1.75. At a=1 area is A(1)=9/4=2.25. But we are interested in a>0 only so it follows that when 0<a1 the area A>1.75.For 1<a<2 The area between x=a and x=2 lies above x-axis whereas that from x=2 and x=a+1 lies below x-axis. But area is a positive quantity

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