Informative line

### Area Under The Curve By Integration

Find the area under the curve by integration. Evaluate the integral by interpreting it in terms of areas. Use the midpoint rule to approximate the definite integrals.

# Area Interpretation to Evaluate the Integral

## Area Interpretation

• Area interpretation means interpreting the area of the region bounded by the curve, axis and limits, using integration.
• Let the curve be $$y=f(x)$$ .

• The shaded portion shown in graph is bounded by

$$\to$$ curve $$y=f(x)$$ at top

$$\to$$ x-axis at bottom

$$\to$$ limit $$x=a$$ on left side

$$\to$$ limit $$x=b$$ on right side

• To calculate the area of this region, we will split the whole region in vertical strips of size $$dx$$.
• The limits of the region are from $$x=a$$ to $$x=b$$, and the curve is defined as $$y=f(x)$$.
• Thus, the area of the shaded region will be

$$\displaystyle A=\int\limits^b_a\;y\;dx$$

Thus, $$\displaystyle A=\int\limits^b_a\;f(x)\;dx$$

Shaded area = Area bounded by $$y=f(x),\;x=a,\;x=b$$ and x-axis.

For example: The graph of a function $$y=f(x)$$ is as  shown. We calculate the area of the shaded region of the function within the limits $$x=0$$ to $$x=2$$ as follows:

Shaded area = Area bounded by $$y=f(x),\;x=a,\;x=b$$ and x-axis

$$\displaystyle A=\int\limits^2_0\;f(x)\;dx$$

$$A=\dfrac{1}{2}(2)(6)$$

$$A=6\;\text{unit}$$

#### The graph of a function $$'f'$$ is as shown, then the value of  $$\displaystyle\int\limits^{10}_{5}f(x)\,dx$$  is

A 49

B $$\dfrac{35}{2}$$

C –10

D 108

×

$$\displaystyle\int\limits^{10}_{5}f(x)\,dx$$ = area below $$y=f(x)$$ and above $$x$$ axis and between $$x=5$$ and $$x=10$$

Shaded area = area of trapezium

$$=\dfrac{1}{2}$$ × sum of parallel sides × height

$$=\dfrac{1}{2}×(2+5)×5$$

$$=\dfrac{35}{2}$$

### The graph of a function $$'f'$$ is as shown, then the value of  $$\displaystyle\int\limits^{10}_{5}f(x)\,dx$$  is

A

49

.

B

$$\dfrac{35}{2}$$

C

–10

D

108

Option B is Correct

# The Midpoint Rule

Mid-point rule

• This rule is used to find the approximate value of the definite integral i.e., the area bounded by curve, axis and limits.
• This rule is also known as rectangle rule as it helps to find the area by dividing it into rectangular strips.
• We can find the area of collection of rectangular strips shown in the figure.
• First, we choose a sample point. (a single observed value, $$x^*_i$$)
• $$x^*_i$$ is mostly taken as right end point of the subinterval because it is convenient. Sometimes, it is chosen as midpoint of the subinterval.
• So, by integral

$$\displaystyle \int\limits^b_a\;f(x)\;dx$$

$$\cong$$ sum of the area of rectangular strips

$$\cong\;\sum\limits^n_{i=1}\;(\text{length}_i)\;(\text{width}_i)$$

$$\cong\;\sum\limits^n_{i=1}\;f(\overline x_i)\;\Delta x$$

The width of each rectangle is $$\Delta x$$.

But length is variable corresponding to the distance of the curve from x-axis.

where, $$\overline x_i=\dfrac{x_{i-1}+x_i}{2}$$

i.e., the mid-point of width of rectangular strip.

• Here, the sample point is chosen as midpoint of the subinterval.

$$\displaystyle \int\limits^b_a\;f(x)\;dx\;\cong\;\sum^n_{i=1}\;f(\overline x_i)\;\Delta x$$

$$=f(\overline x_1)\;\Delta x+f(\overline x_2)\;\Delta x+f(\overline x_3)\;\Delta x+...\;f(\overline x_n)\;\Delta x$$

where $$\Delta x=\dfrac{b-a}{n}$$ is called mid point rule.

• The graph shown represents the sample point as midpoint of the subinterval $$(n=7)$$, interval $$[1,\;8]$$.

For example:

We approximate the integral $$\int\limits^6_1\;2x^2\,dx$$ by the midpoint rule with $$n=5$$ .

$$\displaystyle \int \limits ^6_12x^2\,dx\;\cong\;\sum^5_{i=1}\;f(\overline x_i)\;\Delta x$$

$$\Delta x=\dfrac{6-1}{5}=1,\;f(x)=2x^2,\;\overline x_i=\dfrac{x_{i-1}+x_i}{2}$$

$$\displaystyle\int\limits^6_12x^2\,dx=f(1.5)×1+f(2.5)×1+f(3.5)×1+f(4.5)×1+f(5.5)×1$$

$$=2(1.5)^2+2(2.5)^2+2(3.5)^2+2(4.5)^2+2(5.5)^2$$

$$=2(2.25)+2(6.25)+2(12.25)+2(20.25)+2(30.25)$$

$$=4.5+12.5+24.5+40.5+60.5$$

$$=142.5$$

So, $$\int\limits^6_1\,2x^2\,dx\;\simeq\;142.5$$

#### Approximate the integral $$\displaystyle\int\limits^5_1\dfrac{1}{x^2}dx$$ by the midpoint rule with $$n=4$$.

A 0.7349

B 102

C 0.01

D 0.27

×

$$\displaystyle\int\limits^{5}_{1}\dfrac{1}{x^2}dx\cong \sum\limits^4_{i=1}f(\overline x_i)\Delta x$$

$$\Delta x=\dfrac{5–1}{4}=1,\;f(x)=\dfrac{1}{x^2},\;\overline x_i=\dfrac{x_{i-1}+x_i}{2}$$

$$\displaystyle\int\limits^{5}_{1}\dfrac{1}{x^2}dx=f(1.5)×1+f(2.5)×1+f(3.5)×1+f(4.5)×1$$

$$=\dfrac{1}{(1.5)^2}×1+\dfrac{1}{(2.5)^2}×1+\dfrac{1}{(3.5)^2}×1+\dfrac{1}{(4.5)^2}×1$$

$$=\dfrac{4}{9}+\dfrac{4}{25}+\dfrac{4}{49}+\dfrac{4}{81}$$

$$=0.444+0.16+0.0816+0.0493$$

$$=0.7349$$

$$\displaystyle\int\limits^{5}_{1}\dfrac{1}{x^2}dx\simeq 0.7349$$

### Approximate the integral $$\displaystyle\int\limits^5_1\dfrac{1}{x^2}dx$$ by the midpoint rule with $$n=4$$.

A

0.7349

.

B

102

C

0.01

D

0.27

Option A is Correct

# Property of Integrals

If the greatest and least values of $$f(x)$$ in $$[a,\,b]$$ are $$M,\; m$$ respectively then,

$$\displaystyle m(b–a)\leq\int\limits^{b}_{a}f(x)dx\leq M(b–a)$$

i.e.  $$m\leq f(x)\leq M$$ where $$a\leq x\leq b$$ then

$$\displaystyle m(b–a)\leq\int\limits^{b}_{a}f(x)dx\leq M(b–a)$$

• A function $$f$$ is said to be increasing in an interval $$[a,\,b]$$  if  $$f(x_1)>f(x_2)$$  whenever  $$x_1>x_2$$ for all  $$x_1,\,x_2 \in(a,\,b)$$.

Increasing function

$$x_1>x_2$$

$$\Rightarrow\,f(x_1)>f(x_2)$$

• A function $$f$$ is said to be decreasing in an interval $$[a,\,b]$$  if  $$f(x_1)>f(x_2)$$  whenever  $$x_1<x_2$$ for all  $$x_1,\,x_2 \in(a,\,b)$$.

Decreasing function

$$x_1<x_2$$

$$\Rightarrow\,f(x_1)>f(x_2)$$

• For increasing functions in [a, b], $$f(b)=M$$ and $$f(a)=m$$
• For decreasing function in [a, b], $$f(a)=M,\,f(b)=m$$

For example:

Using above property, we will estimate the value of integral $$\displaystyle\int\limits^4_1\dfrac{1}{x^2}\;dx$$.

Now, it can be seen that

$$f(2)=\dfrac{1}{(2)^2}=\dfrac{1}{4}$$

$$f(3)=\dfrac{1}{3^2}=\dfrac{1}{9}$$

Since, $$\dfrac{1}{4}>\dfrac{1}{9}$$

So, it is a decreasing function of $$x$$ in $$[1,\;4]$$

• Now, we will calculate values of $$M$$ and $$m.$$

$$M=f(1)=1\,\,\,\,\,\,\,,m=f(4)=\dfrac{1}{16}$$

$$\therefore$$By using the property, we have

$$m(b-a)\leq\displaystyle\int\limits^b_a f(x)\;dx\leq M(b-a)$$

$$\dfrac{1}{16}(4-1)\leq\displaystyle\int\limits^4_1\dfrac{1}{x^2}\;dx\leq1(4-1)$$

$$\dfrac{3}{16}\leq\displaystyle\int\limits^4_1\dfrac{1}{x^2}\;dx\leq3$$

So, the values of integral $$I=\displaystyle\int\limits^4_1\dfrac{1}{x^2}\;dx$$ are between $$\dfrac{3}{16}$$ and $$3.$$

#### Estimate the value of $$\displaystyle I=\int\limits^{3}_{0}\dfrac{1}{1+x^3}dx$$.

A $$0\leq I\leq 5$$

B $$1\leq I\leq 7$$

C $$\dfrac{3}{28}\leq I\leq 3$$

D $$–4\leq I\leq \dfrac{2}{7}$$

×

$$f(x)=\dfrac{1}{1+x^3}$$ is a decreasing function of $$x$$ in [0, 3].

$$M=f(0)=1,\,m=f(3)=\dfrac{1}{28}$$

$$\displaystyle\therefore \;\dfrac{1}{28}(3–0)\leq\int\limits^{3}_{0}\dfrac{1}{1+x^3}dx\leq(3–0)×1$$

$$\displaystyle\therefore \;\dfrac{3}{28}\leq\int\limits^{3}_{0}\dfrac{1}{1+x^3}dx\leq3$$

### Estimate the value of $$\displaystyle I=\int\limits^{3}_{0}\dfrac{1}{1+x^3}dx$$.

A

$$0\leq I\leq 5$$

.

B

$$1\leq I\leq 7$$

C

$$\dfrac{3}{28}\leq I\leq 3$$

D

$$–4\leq I\leq \dfrac{2}{7}$$

Option C is Correct

# Evaluating the Integral by Interpreting it in Terms of Area

• If $$f$$ is a positive function then $$\displaystyle \int\limits^{b}_{a}f(x)\,dx$$ = area bounded by the curve $$y=f(x),\;x=a,\;x=b$$ and $$x$$ axis.

Shaded area = $$\displaystyle \int\limits^{b}_{a}f(x)\,dx$$

• If $$f$$ is not a positive function.
• Sometimes, we come across the functions having curve as shown in figure.

• To calculate the total area under the curve, we will calculate A1 and A2 separately.
• Now, we will assign sign to the area A1 and A2.
• Since, A1 is above the x-axis, so it will be positive. Hence, +A1.
• Since, A2 is below the x-axis, so it will be negative. Hence, –A2.
• Negative sign just indicates that the area is below the x-axis.

So, the total area is

$$\displaystyle\int\limits^b_af(x)\,dx=A_1-A_2$$

For example:

We will find the value of $$I=\displaystyle\int\limits^4_{-1}\dfrac{x}{2}\,dx$$ by using area concept.

So, $$I=A_1-A_2$$

$$\text{Area A}_1=\text{Area of}\;\triangle CDE=\dfrac{1}{2}×\text{base × height}$$

$$=\dfrac{1}{2}×(4)(2)$$

$$=4$$

$$\text{Area A}_2=\text{Area of}\;\triangle ABC=\dfrac{1}{2}×(\text{base})\,(\text{height})$$

$$=\dfrac{1}{2}(1)\,\left(\dfrac{1}{2}\right)$$

$$=\dfrac{1}{4}$$

So, $$I=4-\dfrac{1}{4}=\dfrac{15}{4}$$

#### Find the value of $$I=\displaystyle \int\limits^{3}_{–2}(1–2x)\,dx$$ by using area concept.

A 0

B $$\dfrac{7}{2}$$

C –8

D –14

×

$$I=A_1+(–A_2)$$

(area above $$x$$- axis + (– area below $$x$$ - axis))

$$A_1$$= Area of  $$\Delta\, ABC=\dfrac{1}2{}×b×h$$          ($$b$$= base, $$h$$= height)

$$=\dfrac{1}{2}×\dfrac{5}{2}×5$$

$$=\dfrac{25}{4}$$

$$A_2$$ = Area of  $$\Delta\, BDE = \dfrac{1}{2}×b×h$$

$$=\dfrac{1}{2}×\dfrac{5}{2}×5$$

$$=\dfrac{25}{4}$$

$$I=\dfrac{25}{4}+\left(\dfrac{–25}{4}\right)=0$$

### Find the value of $$I=\displaystyle \int\limits^{3}_{–2}(1–2x)\,dx$$ by using area concept.

A

0

.

B

$$\dfrac{7}{2}$$

C

–8

D

–14

Option A is Correct

# Definite Integrals of Piece-wise Defined Functions

For example:

Let $$f(x)=\begin {cases} 10\;\text{if}\;x\leq3\\ –5\;\text{if}\;x>3 \end {cases}$$

then the value of $$\displaystyle\int\limits^{10}_{-5}f(x)\;dx$$ can be calculate as

$$\displaystyle\int\limits^{10}_{-5}f(x)\;dx=\displaystyle\int\limits^{3}_{-5}f(x)\;dx+\displaystyle\int\limits^{10}_{3}f(x)\;dx$$

Here, $$x=3$$ is the value at which $$f$$ is defined as broken.

$$\displaystyle\int\limits^{10}_{-5}f(x)\;dx=\displaystyle\int\limits^{3}_{-5}10\;dx+\displaystyle\int\limits^{10}_{3}-5\;dx$$

$$=10[x]^3_{-5}-5[x]^{10}_{3}$$

$$=10[3-(-5)]-5[10-3]$$

$$=10(8)-5(7)$$

$$=80-35$$

$$=45$$

#### Let  $$f(x)= \begin{cases} 5 & if & x\leq 2\\ –2 & if & x>2\\ \end{cases}$$ then the value of  $$\displaystyle\int\limits^{5}_{–2}f(x)dx$$  is

A 14

B 17

C 82

D –5

×

$$\displaystyle\int\limits^{5}_{–2}f(x)dx=\int\limits^{2}_{–2}f(x)dx\,+\displaystyle\int\limits^{5}_{2}f(x)dx$$

$$x=2$$ is the value of which defined of $$'f'$$ is broken)

$$=\displaystyle\int\limits^{2}_{–2}5\,dx+\int\limits^{5}_{2}–2\,dx$$

$$=5\big[x\big]_{–2}^{\,2}\,–\,2\big[x\big]_2^5$$

$$=5\left(2–(–2)\right)+(–2)(5–2)$$

$$=5×4\;–2×3$$

$$=14$$

### Let  $$f(x)= \begin{cases} 5 & if & x\leq 2\\ –2 & if & x>2\\ \end{cases}$$ then the value of  $$\displaystyle\int\limits^{5}_{–2}f(x)dx$$  is

A

14

.

B

17

C

82

D

–5

Option A is Correct

# Comparison Properties of Integrals

(1) If $$f(x)\geq0$$ in $$[a,\,b]$$ then $$\displaystyle\int\limits^{b}_{a}f(x)dx\geq0$$.

(The value of definite integral of a non negative function is also non negative.)

Example:

If $$2x\geq0$$ in $$[2,\;3]$$, then we will check whether   $$\displaystyle\int\limits^3_22x\;dx \ge0$$

$$\displaystyle\int\limits^3_22x\;dx=2\displaystyle\int\limits^3_2x\;dx$$

$$=2\left[\dfrac{x^2}{2}\right]^3_2$$

$$=\dfrac{2}{2}\left[x^2\right]^3_2$$

$$=9-4$$

$$=5\geq0$$

(2) If $$f(x)\geq g(x)$$ in $$[a,\,b]$$   then $$\displaystyle\int\limits^{b}_{a}f(x)dx\geq\int\limits^{b}_{a}g(x)dx$$.

(A higher graph will have a higher area enclosed between the same limits.)

Example:

We will compare $$\displaystyle\int\limits^2_13x\;dx$$ with $$\displaystyle\int\limits^2_13x^2\;dx$$.

• To compare two integrals with same lower and upper limits, compare the integrand functions.

Now, we know that

$$\begin {array}{c} x^2>x&\text{in}&[1,\;2]\\ 3x^2>3x&\text{in}&[1,\;2] \end {array}|$$

So, $$\displaystyle\int\limits^2_13x^2\;dx>\displaystyle\int\limits^2_13x\;dx$$

Consider all positive values of $$x$$, then, in general  $$x^n>x^m$$  when $$n>m$$ and $$x>1$$ and $$x^n<x^m$$ when $$n>m$$ and $$x\in(0,\,1)$$

#### Choose the correct option.

A $$\displaystyle\int\limits^{1}_{0}2x^2dx<\int\limits^{1}_{0}2x^3dx$$

B $$\displaystyle\int\limits^{5}_{2}2x^2dx<\int\limits^{5}_{2}2x^3dx$$

C $$\displaystyle\int\limits^{1}_{0}x^4dx<\int\limits^{1}_{0}x^5dx$$

D $$\displaystyle\int\limits^{3}_{1}x^4dx<\int\limits^{3}_{1}x^3dx$$

×

To compare the two integrals with same lower and upper limits, compare the integrand functions.

$$x^2>x^3$$ in $$[0,\,1]$$

$$\Rightarrow\,2x^2>2x^3$$ in $$[0,\,1]$$

$$\therefore\,$$ option 'a' is false.

$$x^2<x^3$$ in $$[2,\,5]$$

$$\Rightarrow\,2x^2<2x^3$$ in $$[2,\,5]$$

$$\therefore\,$$ option 'b' is true.

$$x^4>x^5$$ in $$[0,\,1]$$

$$\therefore\,$$ option 'c' is false.

$$x^4>x^3$$ in $$[1,\,3]$$

$$\therefore\,$$ option 'd' is false.

### Choose the correct option.

A

$$\displaystyle\int\limits^{1}_{0}2x^2dx<\int\limits^{1}_{0}2x^3dx$$

.

B

$$\displaystyle\int\limits^{5}_{2}2x^2dx<\int\limits^{5}_{2}2x^3dx$$

C

$$\displaystyle\int\limits^{1}_{0}x^4dx<\int\limits^{1}_{0}x^5dx$$

D

$$\displaystyle\int\limits^{3}_{1}x^4dx<\int\limits^{3}_{1}x^3dx$$

Option B is Correct