#### To determine

**a) **

**To evaluate:**

The Riemann sum for fx

#### Answer

1.25

Riemann sum using right endpoints for the sample points is an over estimate of the area under the graph of fx=x2-x

#### Explanation

**Concept:**

Definition of Riemann sum:

∑i=1nfxi* ∆x

is called as Riemann sum

Where, ∆x= b-an, xi=a+i ∆x, &xi*∈xi-1, xi

Riemann sum is the sum, of the areas of the rectangles that lie above the *x*-axis and the negative of the areas of the rectangles that lie below the *x*-axis

**Given:**

fx=x2-x, 0≤x≤2

**Calculation:**

First, write the expression of the Riemann sum

Four subintervals are given, that means, n=4

Taking the sample points as the right endpoints means that

The right endpoints are 0.5, 1, 1.5, 2

By definition of Riemann sum

∑i=1nfxi ∆x=∑i=14fxi ∆x

So now calculate ∆x,

∆x= b-an=2-04=12

Substitute this value in Riemann sum expression,

=∑i=14fxi12

=∑i=14xi2-xi12

=12f0.5+f1+f1.5+f2

=120.52-0.5+12-1+1.52-1.5+22-2

=12-0.25+0+0.75+2

=122.5

=1.25

Draw the diagram of Riemann sum for right endpoints

Riemann sum using right endpoints for the sample points is an over estimate of the area under the graph of fx=x2-x

**Conclusion:**

Therefore,

∑i=1nfxi ∆x≈1.25

Riemann sum using right endpoints for the sample points is an over estimate of the area under the graph of fx=x2-x

#### To determine

**b) **

**To calculate:**

The value of the integral ∫02x2-xdx

#### Answer

23

#### Explanation

**Concept:**

Definition of Definite integral:

∫abfxdx=limn→∞∑i=1nfxi ∆x

Where,

∆x= b-an and xi=a+i ∆x

**Formula:**

∑i=1ni2=nn+12n+16

∑i=1ni=nn+12

**Calculation:**

Let

∫abx2-xdx=limn→∞∑i=1nxi2-xi ∆x

Find out ∆x &xi,

∆x= b-an

= 2-0n

∆x= 2n

xi=a+i ∆x

=0+i·2n

xi=2in

Substitute the values of ∆x &xi in evaluating the limit

∫02x2-xdx=limn→∞∑i=1n2in2-2in2n

=limn→∞2n∑i=1n4i2n2-2in

=limn→∞2n∑i=1n4i2n2-∑i=1n2in

=limn→∞2n4n2∑i=1ni2-2n∑i=1ni

Substituting for the sums,

=limn→∞2n4n2·nn+12n+16-2n·nn+12

=2limn→∞2n2·n+12n+13-1n·n+1

=2limn→∞23·n+12n+1n·n-1n·n+1

=2limn→∞23·n+1n·2n+1n-n+1n

=2limn→∞23·1+1n·2+1n-1+1n

Simplify the limit,

=223·1+0·2+0-1+0

=243-1

=23

**Conclusion:**

Therefore,

∫02x2-xdx=23

#### To determine

**c) **

**To check:**

∫02x2-xdx=23

#### Answer

23

#### Explanation

**Concept:**

The Fundamental Theorem of Calculus-Suppose f is continuous on [a, b] then

∫abfxdx=Fb-F(a),F is an Antiderivative of f i.e F'=f

**Calculation:**

Consider, fx=x2-x

Take antiderivative, and by using power rule of integration,

Fx=x33-x22

By Fundamental Theorem of Calculus,

∫02x2-xdx=F2-F0

=233-222-033+022

=83-42

=23

**Conclusion:**

Therefore, By Fundamental Theorem of Calculus,

∫02x2-xdx=23

#### To determine

**d) **

**To draw:**

Graph of the function fx=x2-x

#### Answer

#### Explanation

**Concept:**

The net area under the curve is the area under the curve under the x-axis subtracted from the area under the curve above x-axis

**Calculation:**

Graph of the function fx=x2-x is

The net area under the curve is

∫02x2-xdx=A2-A1

**Conclusion:**

Therefore, by graph of the function fx=x2-x

The net area under the curve is

∫02x2-xdx=A2-A1