#### To determine

a)

**To describe:**

A relation

#### Answer

*Ln≤A≤Rn*

#### Explanation

**1) Concept:**

The area *A* of the region *S* that lies under the graph of the continuous function *f* is the limit of the sum of the areas of approximating rectangles.

*A=limn→∞Rn=limn→∞fx1∆x+fx2∆x+…+fxn∆x=limn→∞∑i=1nfxi∆x*

The width of the interval *a, b* is *(b-a)*, so the width of each *n* strip is

*∆x=b-an*

where *x0=a * and * xn=b *. The right end points of the subintervals are *xi=a+i∆x.*

2) **Calculation:**

It is given that A is the area under the graph of an increasing continuous function from *a to b*.

Also, *Ln* is the approximation to *A* using the left end point of the interval, and *Rn* is the approximation to *A* using the right end point of the interval.

So, if the function is increasing and continuous, then the left approximation is less than the actual area and the right approximation is larger than the actual area.

That means

*Ln≤A≤Rn*

**Conclusion:**

*Ln≤A≤Rn*

#### To determine

b)

**To show:**

*Rn-Ln=b-anfb-fa*

#### Answer

*Rn-Ln=b-anfb-fa*

#### Explanation

1) **Calculation:**

The equation *Rn* means *∆x* times the sum from * x1 to xn*, where * ∆x=b-an*

So, equation *Rn* is

*Rn=∑i=1nfxi∆x*

And *xi=a+i∆x*

Substitute *xi=a+b-ain* in *Rn*.

*Rn=∑i=1nfa+b-ainb-an*

Similarly, equation *Ln* means *∆x* times the sum from * x0 to xn-1*, where * ∆x=b-an*

So, equation *Ln* is

*Ln=∑i=1nfxi∆x*

We have *xi=a+i∆x*.

Substitute *xi=a+b-ain* in * Ln*.

*Ln=∑i=0n-1fa+b-ainb-an*

Subtract the *Ln* from *Rn;* we get

*Rn-Ln=∑i=1nfa+b-ainb-an-∑i=0n-1fa+b-ainb-an*

*=b-an∑i=1nfa+b-ain-∑i=0n-1fa+b-ain*

*=b-an∑i=1n-1fa+b-ain+fa+b-ann-fa+b-a0n-∑i=1n-1fa+b-ain*

*=b-anfa+b-a-fa*

*=b-anfb-fa*

Write the whole equation.

*Rn-Ln=b-anfb-fa*

This is the required equation

**Conclusion:**

*Rn-Ln=b-anfb-fa*

#### To determine

c)

**To deduce:**

Rn-A<b-anfb-fa

#### Answer

*Rn-A<b-anfb-fa*

#### Explanation

1) **Calculation:**

From part a),

*Ln≤A≤Rn*

Take

*Ln<A*

Add *Rn* on both the sides.

*Ln+Rn<A+Rn*

By subtracting *A & Ln* from both sides,

*Rn-A<Rn-Ln*

But from part b),

*Rn-Ln=b-anfb-fa*

So substitute this equality in * Rn-A<Rn-Ln*.

*Rn-A<b-anfb-fa*

**Conclusion:**

*Rn-A<b-anfb-fa*