Informative line

### Approximate Area Under The Graph

Find the area under the graph & a curve approximate the area using rectangles. Practice upper and lower estimates, area under a curve calculus by rectangular approximation method.

# The Area Problem

From the earlier classes we have learned the formulas for finding the areas of some figures, which have only straight lines as sides. Some of these are given below.

(1)

(2)

(3)

(4)

#### Consider a rectangle whose length and breadth are $$5\,cm$$ and $$7\,cm$$ respectively and a square whose side is $$6\,cm$$. If $$R$$ is the area of rectangle and $$S$$ is the area of square then,

A $$R>S$$

B $$R<S$$

C $$R=S$$

D $$R=2S$$

×

$$R=\ell×b$$

$$=5cm×7cm$$

$$=35\,\text{cm}^2$$

$$S=a^2$$

$$=(6cm)^2$$

$$=36\,\text{cm}^2$$

$$R<S$$

### Consider a rectangle whose length and breadth are $$5\,cm$$ and $$7\,cm$$ respectively and a square whose side is $$6\,cm$$. If $$R$$ is the area of rectangle and $$S$$ is the area of square then,

A

$$R>S$$

.

B

$$R<S$$

C

$$R=S$$

D

$$R=2S$$

Option B is Correct

# The Area of Regions which have Curved Sides

It is not so easy to find the area of regions which have curved sides. We first start by approximating the area bounded by curved sides through rectangles.

Consider the curve $$\to\;y=x^3$$

Suppose we want to estimate the area under $$y=x^3$$ from 0 to 1, base being $$x$$ axis (shaded portion).

We can easily say that this area is less than that of square of length 1 so, shaded area < 1.

Since, area is a positive quantity, we say that $$0<A<1$$.

#### Let $$A$$ be the area under the curve $$y=x^4$$ from 0 to 1, base being $$x$$ axis, as shown by shaded region in the given figure, then

A $$A>5$$

B $$0<A<1$$

C $$A>7$$

D $$A>2$$

×

Area is less than square of side length 1

Area is a physical quantity so, it is always positive.

$$\therefore\;0<A<1$$

### Let $$A$$ be the area under the curve $$y=x^4$$ from 0 to 1, base being $$x$$ axis, as shown by shaded region in the given figure, then

A

$$A>5$$

.

B

$$0<A<1$$

C

$$A>7$$

D

$$A>2$$

Option B is Correct

# Estimating the Lower Bound of Area by using  Left End Points

If we use the left end point height of the rectangles, we get a value which is a  lower estimate of the area.

Consider  $$y=x^3$$  in [0,1].

We will use four  rectangles.The area we calculate is shaded and is obviously less than the actual area.

$$L_4 = \dfrac{1}{4} × 0+\dfrac{1}{4}× \left(\dfrac{1}{4}\right)^3+\dfrac{1}{4}× \left(\dfrac{1}{2}\right)^3+\dfrac{1}{4}× \left(\dfrac{3}{4}\right)^3$$

(The left height of first rectangle is 0)

$$= \dfrac{1}{4} \left[\dfrac{1}{64}+\dfrac{1}{8}+\dfrac{27}{64}\right]$$

$$= \dfrac{1}{4}\left[\dfrac{1+8+27}{64}\right]$$

$$= \dfrac{9}{64}$$

$$=0 .1406$$

We say the area $$A>0 .1406$$

$$L_n$$ will represent the area obtained by dividing the interval into n equal parts  and using left end points for heights.

$$R_n$$ = Shaded area

$$R_n>A$$

$$L_n$$= Shaded area

$$L_n<A$$

• As n increases, the values of $$R_n$$ and $$L_n$$ will approach the actual area.
• If the function is decreasing, we will observe that  $$R_n<A$$  and  $$L_n >A$$ .

#### Estimate the lower bound of area bounded by $$y=\sqrt x$$  in [0,1] by calculating $$L_4$$ .

A $$A>0.518$$

B $$A=7$$

C $$A=-2$$

D $$A=11$$

×

Divide the interval $$[0,1]$$ into four equal parts i.e., $$\left[0,\dfrac{1}{4}\right],\left[\dfrac{1}{4},\dfrac{1}{2}\right],\left[\dfrac{1}{2},\dfrac{3}{4}\right],\left[\dfrac{3}{4},1\right]$$

$$L_4 = \dfrac{1}{4} × \sqrt 0 + \dfrac{1}{4}× \sqrt{\dfrac{1}{4}}+ \dfrac{1}{4}× \sqrt{\dfrac{1}{2}} + \dfrac{1}{4}× \sqrt{\dfrac{3}{4}}$$

$$= \dfrac{1}{4}× \left[\dfrac{1}{2}+\dfrac{1}{\sqrt2} +\dfrac{\sqrt3}{2} \right]$$

$$= \dfrac{1}{4}× \left[.5+.707 +.866 \right]$$

$$=0.518$$

$$\Rightarrow A>0.518$$

### Estimate the lower bound of area bounded by $$y=\sqrt x$$  in [0,1] by calculating $$L_4$$ .

A

$$A>0.518$$

.

B

$$A=7$$

C

$$A=-2$$

D

$$A=11$$

Option A is Correct

# Estimating the Areas by using Left End Points and Right End Points

For an increasing function, we can get a closer estimate to the areas bounded by curved sides by breaking it into parts and then estimating each part by a rectangle.

Consider $$y=x^3$$ in $$[0,\,1]$$

We divide $$[0,\,1]$$ into two parts $$\to\;\left[0,\,\dfrac{1}{2}\right]$$ and $$\left[\dfrac{1}{2},\,1\right]$$. Now, we observe that shaded area can be approximated by sum of two rectangles.

$$\therefore$$ Area of rectangles $$=\dfrac{1}{2}×\left(\dfrac{1}{2}\right)^3+\dfrac{1}{2}×1^3$$

$$=\dfrac{1}{2}×\dfrac{1}{8}+\dfrac{1}{2}×1$$

$$=\dfrac{1}{16}+\dfrac{1}{2}$$

$$=\dfrac{9}{16}=0.5625$$

$$\therefore$$ Shaded Area $$=A<0.56$$

We have obtained a closer estimate to the area. We can get even better approximation by increasing the number of rectangles and ultimately an accurate area value by taking the limit.

• In this approximation, we have used the right end points as heights.

### Estimating the Lower Bound of Area by Using  Left End Points

If we use the left end point height of the  rectangles we get a value which is a  lower  estimate of the area.

Consider $$y=x^3$$ in [0,1].

We will use four  rectangles. The area we calculate is shaded and is obviously less than the actual area.

$$L_4 = \dfrac{1}{4} × 0+\dfrac{1}{4}× \left(\dfrac{1}{4}\right)^3+\dfrac{1}{4}× \left(\dfrac{1}{2}\right)^3+\dfrac{1}{4}× \left(\dfrac{3}{4}\right)^3$$

(The left height of first rectangle is 0)

$$= \dfrac{1}{4} \left[\dfrac{1}{64}+\dfrac{1}{8}+\dfrac{27}{64}\right]$$

$$= \dfrac{1}{4}\left[\dfrac{1+8+27}{64}\right]$$

$$= \dfrac{9}{64}$$

$$= .1406$$

We say the area $$A> .1406$$

$$L_n$$ will represent the area obtained by dividing the interval into n equal parts  and using left end points for heights.

$$R_n$$= Shaded Area

$$R_n>A$$

$$L_n$$= Shaded Area

$$L_n <A$$

• As n increases the values of $$R_n$$ and $$L_n$$ will approach the actual area.
• If the function is decreasing,  we will observe that $$R_n<A$$ and $$L_n >A$$ .

#### Using four rectangles, find lower and upper estimate  of area under the given  graph of a function $$'f'$$ from $$x = 0$$ to $$x = 8$$ .

A $$32<A<46$$

B $$A>49$$

C $$2<A<4$$

D $$-1<A<18$$

×

Divide the interval [0,8] into 4 equal parts  i.e.,  [0,2], [2,4], [4,6], [6,8].

$$R_4 = 2×\underline 3 + 2 ×\underline 5 +2× \underline7 +2×\underline8$$

$$= 2[3+5+7+8]$$

$$=2×23$$

$$=46$$

The underlined values are $$f(2), f(4), f(6) \,\,{\text{&}}\,\, f(8)$$.

$$R_4 = 46$$

$$L_4 = 2× 1+2×3+2× 5+2×7$$

$$=2[1+3+5+7]$$

$$= 2× 16$$

$$= 32$$

$$L_4 =32$$

$$32<A<46$$

### Using four rectangles, find lower and upper estimate  of area under the given  graph of a function $$'f'$$ from $$x = 0$$ to $$x = 8$$ .

A

$$32<A<46$$

.

B

$$A>49$$

C

$$2<A<4$$

D

$$-1<A<18$$

Option A is Correct

# Estimation to the Areas Bounded by Curved Sides by Breaking it into Parts

We can get a closer estimate to the areas bounded by curved sides by breaking it into parts and then estimating each part by a rectangle.

Consider $$y=x^3$$ in $$[0,\,1]$$

We divide $$[0,\,1]$$ into two parts $$\to\;\left[0,\,\dfrac{1}{2}\right]$$ and $$\left[\dfrac{1}{2},\,1\right]$$

Now, we observe that shaded area can be approximated by sum of two rectangles

$$\therefore$$ Area of rectangles $$=\dfrac{1}{2}×\left(\dfrac{1}{2}\right)^3+\dfrac{1}{2}×1^3$$

$$=\dfrac{1}{2}×\dfrac{1}{8}+\dfrac{1}{2}×1$$

$$=\dfrac{1}{16}+\dfrac{1}{2}$$

$$=\dfrac{9}{16}=0.5625$$

$$\therefore$$ Shaded Area $$=A<0.56$$

We have obtained a closer estimate to the area. We can get even better approximation by increasing the number of rectangles and ultimately an accurate area value by taking the limits.

• In this approximation, we have used the right end points as heights.

#### The estimated value of area bounded by $$y=x^2$$  from 0 to 1 by using two rectangles is

A $$A>2$$

B $$A<0.625$$

C $$A>-2$$

D $$A<0.1$$

×

Divide the interval $$[0,\,1]$$ into two equal parts $$\to\;\left[0,\,\dfrac{1}{2}\right]$$ and $$\left[\dfrac{1}{2},\,1\right]$$

Now from figure,

$$R_2=$$ sum of areas of two rectangles

$$=\dfrac{1}{2}×\dfrac{1}{4}+\dfrac{1}{2}×1$$

$$=\dfrac{1}{8}+\dfrac{1}{2}$$

$$=\dfrac{5}{8}=0.625$$

$$\therefore\;A<0.625$$

$$R_n$$ now becomes the symbol for sum of area of $$n$$ rectangles formed by dividing the interval into $$n$$ parts and taking right heights or right end points.

### The estimated value of area bounded by $$y=x^2$$  from 0 to 1 by using two rectangles is

A

$$A>2$$

.

B

$$A<0.625$$

C

$$A>-2$$

D

$$A<0.1$$

Option B is Correct

# Approximating the Area using Four Rectangles (R4)

We can get a closer estimate to the area bounded by curved sides by breaking it into parts and then estimating end part by a rectangle.

Consider  $$y=x^6$$ in $$[0,\,1]$$

The area bounded is as shown.

Now this area can be approximated by dividing $$[0,\,1]$$ into four equal intervals $$\left[0,\,\dfrac{1}{4}\right],\;\left[\dfrac{1}{4},\,\dfrac{1}{2}\right],\;\left[\dfrac{1}{2},\,\dfrac{3}{4}\right]$$ and $$\left[\dfrac{3}{4},\,1\right]$$

and calculating the area of rectangle formed.

$$(A_1,\,A_2,\,A_3,\;\&\;\,A_4)$$

$$R_4=$$ estimated area using four rectangles

$$=\dfrac{1}{4}×\underbrace{\left(\dfrac{1}{4}\right)^6}_{A_1}+\dfrac{1}{4}× \underbrace{\left(\dfrac{1}{2}\right)^6}_{A_2}+\dfrac{1}{4}×\underbrace{\left(\dfrac{3}{4}\right)^6}_{A_3}+\underbrace{\dfrac{1}{4}×1^6}_{A_4}$$

• This value will be larger than shaded area which was to be found.

$$\therefore$$ We say that $$A<R_4$$.

#### Estimate the area A bounded by the curve $$y=x^3$$ in $$[0,\,1]$$ by taking four rectangles i.e., $$R_4$$ .

A $$A<0.3906$$

B $$A>5$$

C $$A=7$$

D $$A<-2$$

×

Divide the interval $$[0,\,1]$$ into four equal parts

i.e., $$\left[0,\,\dfrac{1}{4}\right],\;\left[\dfrac{1}{4},\,\dfrac{1}{2}\right],\;\left[\dfrac{1}{2},\,\dfrac{3}{4}\right]$$ and $$\left[\dfrac{3}{4},\,1\right]$$ .

$$R_4=\dfrac{1}{4}×\left(\dfrac{1}{4}\right)^3+\dfrac{1}{4}×\left(\dfrac{1}{2}\right)^3+\dfrac{1}{4}×\left(\dfrac{3}{4}\right)^3+\dfrac{1}{4}×1^3$$

$$=\dfrac{1}{4}×\dfrac{1}{64}+\dfrac{1}{4}×\dfrac{1}{8}+\dfrac{1}{4}×\dfrac{27}{64}+\dfrac{1}{4}$$

$$=\dfrac{100}{256}$$

$$=0.3906$$

$$A<0.3906$$

### Estimate the area A bounded by the curve $$y=x^3$$ in $$[0,\,1]$$ by taking four rectangles i.e., $$R_4$$ .

A

$$A<0.3906$$

.

B

$$A>5$$

C

$$A=7$$

D

$$A<-2$$

Option A is Correct

# Estimating the Area for a Decreasing Function By Finding $$R_n$$  and $$L_n$$

If $$f$$ is  a decreasing function , then $$R_n$$ will give the lower estimate and $$L_n$$ will give the upper estimate of the actual area.

Estimation using $$R_n=Shaded\;Area$$

Estimation using $$L_n=Shaded\; Area$$

$$R_n<A<L_n$$

#### Using five rectangles, find the upper and lower estimate of the area under the given graph of $$'f'$$ from $$x =0$$ to $$x = 10$$ .

A $$54<A<72$$

B $$A>96$$

C $$A<34$$

D $$2<A<3$$

×

Divide the interval [0,10] into five equal parts, i.e., $$[0,2], [2,4],[4,6],[6,8],[8,10]$$.

$$R_5 = 2× f(2) + 2× f(4) + 2×f(6) +2× f(8) +2×f(10)$$

$$= 2× 9+2× 8+2× 6+2× 3+2× 1$$

$$=2[9+8+6+3+1]$$

$$\;= 2× 27\\ = 54$$

$$\Rightarrow R_5 =54$$

$$\Rightarrow L_5 = 2× f(0) +2×f(2) + 2× f(4) +2× f(6)+2× f(8)$$

$$= 2× 10+2× 9+2× 8+2× 6+2× 3$$

$$= 2[10+9+8+6+3]$$

$$\;=2× 36\\=72$$

$$\Rightarrow L_5 = 72$$

$$54<A<72$$

### Using five rectangles, find the upper and lower estimate of the area under the given graph of $$'f'$$ from $$x =0$$ to $$x = 10$$ .

A

$$54<A<72$$

.

B

$$A>96$$

C

$$A<34$$

D

$$2<A<3$$

Option A is Correct

# Approximating Area of Regions Bounded by Functions Whose Expressions are Known

Suppose instead of graphs we are given an expression of the function. We sketch that (using standard graphs and graphical transformation) and then approximate using $$R_n \,\,{\text{or}}\,\, L_n$$ as the case may be

• If a function is not increasing or decreasing in an interval there is no fixed relation between $$R_n,\,L_n \,\,{\text {and}}\,\,A$$.

#### The area bounded by $$f(x)= 3+sin\,x$$  and $$x- axis$$ between $$x =0$$ and $$x = \pi$$ can be approximated as (use four rectangles, $$R_4\,\,\text{and} \,\,L_4$$)

A $$A\cong11.315$$

B $$A\cong52$$

C $$A\cong-\dfrac{1}{2}$$

D $$A\cong.01$$

×

Difficult to judge whether shaded area is more or less than required area.

Sketch $$y=sin\,x$$ and apply the transformation

$$y= f(x) +C$$ where $$C=3$$   .

Divide the interval  $$[0,\pi]$$ into four equal parts i.e., $$\left(0,\dfrac{\pi}{4}\right),\left(\dfrac{\pi}{4},\dfrac{\pi}{2}\right),\left(\dfrac{\pi}{2},\dfrac{3\pi}{4}\right),\left(\dfrac{3\pi}{4},\pi\right)$$

$$R_4 = \dfrac{\pi}{4}× f\left(\dfrac{\pi}{4}\right) +\dfrac{\pi}{4}× f\left(\dfrac{\pi}{2}\right) + \dfrac{\pi}{4}× f\left(\dfrac{3\pi}{4}\right) +\dfrac{\pi}{4}× f\left(\pi\right)$$

$$= \dfrac{\pi}{4}×\Big[\left(3+\dfrac{1}{\sqrt{2}}\right)+4+\left(3+\dfrac{1}{\sqrt{2}}\right)+3\Big]$$

$$=\dfrac{\pi}{4}[14.414]$$

$$=11.315$$

$$L_4 = \dfrac{\pi}{4}×f(0)+\dfrac{\pi}{4}× f\left(\dfrac{\pi}{4}\right) +\dfrac{\pi}{4}× f\left(\dfrac{\pi}{2}\right) + \dfrac{\pi}{4}× f\left(\dfrac{3\pi}{4}\right)$$$$=\dfrac{\pi}{4}× \left[3+\left(3+\dfrac{1}{\sqrt2}\right)+4+\left(3+\dfrac{1}{\sqrt2}\right) \right]$$

$$=\dfrac{\pi}{4} [14.414] =11.315$$

We observe that,  $$R_4 = L_4 = 11.315$$

We say, $$A\cong11.315$$

### The area bounded by $$f(x)= 3+sin\,x$$  and $$x- axis$$ between $$x =0$$ and $$x = \pi$$ can be approximated as (use four rectangles, $$R_4\,\,\text{and} \,\,L_4$$)

A

$$A\cong11.315$$

.

B

$$A\cong52$$

C

$$A\cong-\dfrac{1}{2}$$

D

$$A\cong.01$$

Option A is Correct