Informative line

### Definite Integration

Learn definition of the definite integral with all the steps. Evaluate definite integration as a limit of riemann Sum & sigma notation calculus formula.

# The Definition of Definite Integral

Let  $$f$$ be a function defined in [a, b], we divide the interval [a, b] into $$n$$ subintervals; whose width is

$$\Delta x=\dfrac{b-a}{n}$$

Let $$x_0(a), \;x_1,\,x_2,......x_{n-1},\;x_n(b)$$ be the end points of the subintervals $$[x_0,\;x_1],\;[x_1,\;x_2],.......[x_{n-1,\;x_n}]$$. Let $$x_1^*,\;x_2^*.......x_n^*$$ be sample points in these subintervals such that $$x_1^*\in\;\;[x_{i-1},\;x_i]$$.

• The definite integral of $$'f'$$ from $$'a'$$  to $$'b'$$ is defined as

$$\displaystyle\int \limits^b_a f(x)dx=\lim\limits _{n\to\infty}\sum^n_{i=1}(f(x_i^*)\Delta x)$$

• This assumes that the limit exists. If this limit can be found we say $$'f'$$ integrable in [a, b].
• $$'a'$$ is called lower limit, $$'b'$$ the upper limit of integral.
• $$'f'$$ is called integral function and process of calculating integral is called integration.

#### Express the limit as a definite integral on given interval. $$\lim\limits _{n\to\infty\,\,\,}\sum\limits^n_{i=1}\left(\dfrac{2+x_i}{5+3x_i}\right)\Delta x$$, $$x\in[5,\,9]$$

A $$\displaystyle\int \limits^{13}_{11} \dfrac{2-x}{5+x}dx$$

B $$\displaystyle\int \limits^{9}_{5} \left(\dfrac{2+x}{5+3x}\right)dx$$

C $$\displaystyle\int \limits^{12}_{7} \dfrac{2x+1}{x-2}dx$$

D $$\displaystyle\int \limits^{50}_{7} x\;dx$$

×

$$\displaystyle\int \limits^b_a f(x)dx=\lim\limits _{n\to\infty}\,\,\sum^n_{i=1}\left(f(x_i^*)\Delta x\right)\,\,\,\,\to x_i^*\text{ can be} \,x_i\,or\;x_{i-1}$$.

Here,  $$a=5,\;b=9,\;f(x_i)=\dfrac{2+x_i}{5+3x_i}$$

$$\therefore\,\lim\limits _{n\to\infty}\,\,\sum\limits^n_{i=1}\left(\dfrac{2+x_i}{5+3x_i}\right)\Delta x$$        ( Replace $$\left(\lim\sum\right)$$ by $$\int$$)

$$=\displaystyle\int \limits^{9}_{5} \left(\dfrac{2+x}{5+3x}\right)dx$$          ($$\Delta x$$ is replaced by $$dx$$)

### Express the limit as a definite integral on given interval. $$\lim\limits _{n\to\infty\,\,\,}\sum\limits^n_{i=1}\left(\dfrac{2+x_i}{5+3x_i}\right)\Delta x$$, $$x\in[5,\,9]$$

A

$$\displaystyle\int \limits^{13}_{11} \dfrac{2-x}{5+x}dx$$

.

B

$$\displaystyle\int \limits^{9}_{5} \left(\dfrac{2+x}{5+3x}\right)dx$$

C

$$\displaystyle\int \limits^{12}_{7} \dfrac{2x+1}{x-2}dx$$

D

$$\displaystyle\int \limits^{50}_{7} x\;dx$$

Option B is Correct

# Independency of Variable in Definite Integration

• $$\displaystyle\int\limits^b_af(x)dx=\displaystyle\int\limits^b_af(t)dt=\int\limits^b_af(z)dz$$

The definite integral does not depend upon variable of integration $$x$$, we can replace $$x$$ by $$t$$ or $$z$$, without changing the function $$'f'$$.

#### If  $$\displaystyle\int\limits^\pi_0(sin\,x)dx=2$$  then the value of  $$\displaystyle\int\limits^\pi_0(sin\,t)dt$$  is

A 2

B –2

C 18

D –44

×

$$\displaystyle\int\limits^b_af(x)dx=\displaystyle\int\limits^b_af(t)dt$$

$$\displaystyle\int\limits^\pi_0(sin\,x)dx=\int\limits^\pi_0(sin\,t)dt$$         (Only $$x$$ is replaced by $$'t'$$)

= 2

### If  $$\displaystyle\int\limits^\pi_0(sin\,x)dx=2$$  then the value of  $$\displaystyle\int\limits^\pi_0(sin\,t)dt$$  is

A

2

.

B

–2

C

18

D

–44

Option A is Correct

# The Riemann Sum and Definite Integrals  • $$\displaystyle\int\limits^b_af(x)dx=\lim\limits_{n\to \infty}\,\left(\sum\limits^n_{i=1}\,f(x_i)\Delta x\right)$$

where $$x_i$$ is the right end point (Normally we use $$x_i^*=x_i$$ for convenience)

• $$\Delta x=\dfrac{b–a}{n},\;x_i=a+i\Delta x$$
• Express the entire expression $$E$$ in terms of $$x$$ and then take the limit.

#### Evaluate $$\displaystyle\int\limits^4_2(x^2+x)dx$$ by calculating the limit of the sum.

A $$\dfrac{74}{3}$$

B 1

C –18

D $$\dfrac{8}{9}$$

×

$$\displaystyle\int\limits^4_2(x^2+x)dx=\lim\limits_{n\to \infty}\,\sum\limits^n_{i=1}\,f(x_i)\Delta x$$

$$\Delta x=\dfrac{4–2}{n}=\dfrac{2}{n},\;\;x_i=2+\dfrac{2i}{n}$$

$$\displaystyle\int\limits^4_2(x^2+x)dx=\lim\limits_{n\to \infty}\,\sum\limits^n_{i=1}\,f\left(2+\dfrac{2i}{n}\right)×\dfrac{2}{n}$$

$$=\lim\limits_{n\to \infty}\,\sum\limits^n_{i=1}\,\left[\left(2+\dfrac{2i}{n}\right)^2+\left(2+\dfrac{2i}{n}\right)\right]×\dfrac{2}{n}$$

$$=\lim\limits_{n\to \infty}\,\sum\limits^n_{i=1}\,\left(4+\dfrac{4i^2}{(n)^2}+\dfrac{8i}{n}+2+\dfrac{2i}{n}\right)×\dfrac{2}{n}$$

$$=\lim\limits_{n\to \infty}\,\left(\sum\limits^n_{i=1}\,4+\dfrac{4}{n^2}\sum\limits^n_{i=1} i^2+\dfrac{8}{n}\sum\limits^n_{i=1} i+\sum\limits^n_{i=1}2+\dfrac{2\sum\limits^n_{i=1} i}{n}\right)×\dfrac{2}{n}$$

$$=\lim\limits_{n\to \infty}\left(4n+\dfrac{4n(n+1)(2n+1)}{6n^2}+\dfrac{8n(n+1)}{2n}+2n+\dfrac{2n(n+1)}{2n}\right)×\dfrac{2}{n}$$

$$=\lim\limits_{n\to \infty}\left(8+\dfrac{4}{3}\dfrac{(n+1)(2n+1)}{n^2}+\dfrac{8(n+1)}{n}+4+\dfrac{2(n+1)}{n}\right)$$

$$=\lim\limits_{n\to \infty}\left[8+\dfrac{4}{3}\left(\left(1+\dfrac{1}{n}\right)\left(2+\dfrac{1}{n}\right)\right)+8\left(1+\dfrac{1}{n}\right)+4+2\left(1+\dfrac{1}{n}\right)\right]$$

$$=8+\dfrac{8}{3}+8+4+2$$        $$\left(n\to \infty \Rightarrow\,\dfrac{1}{n}\to 0\right)$$

$$=\dfrac{74}{3}$$

### Evaluate $$\displaystyle\int\limits^4_2(x^2+x)dx$$ by calculating the limit of the sum.

A

$$\dfrac{74}{3}$$

.

B

1

C

–18

D

$$\dfrac{8}{9}$$

Option A is Correct

#### Evaluate  $$\sum\limits^{20}_{i=1}\;(2i^2+3i)$$ .

A 5000

B 6370

C 12

D –89

×

$$\sum\limits^{20}_{i=1}\;(2i^2+3i)$$ $$=\sum\limits^{20}_{i=1}\;2i^2+\sum\limits^{20}_{i=1}\;3i$$

$$\bigg(\sum (a_i+b_i)=\sum a_i+\sum b_i\bigg)$$

$$=2\sum\limits^{20}_{i=1}\;i^2+3\sum\limits^{20}_{i=1}\;i$$        $$\bigg(\sum c_ia=a\sum c_i\bigg)$$

$$=\dfrac{20n(n+1)(2n+1)}{6}+\dfrac{3n(n+1)}{2}$$ with $$n=20$$

$$=2×\dfrac{20×21×41}{6}+\dfrac{3×20×21}{2}$$

$$=5740+630$$

$$=6370$$

### Evaluate  $$\sum\limits^{20}_{i=1}\;(2i^2+3i)$$ .

A

5000

.

B

6370

C

12

D

–89

Option B is Correct

# Riemann Sum

The sum  $$\sum\limits^n_{i=1}\,f(x_i^*)\,\Delta x$$ where $$\Delta x=\dfrac{b–a}{n}$$ is called the Riemann Sum,

after a German  Mathematician Bernhard Riemann.

• If $$'f'$$ is a positive function then Riemann sum is sum of areas of rectangles which approximate the area bounded by a positive function.
• If $$f(x)\geq0$$  $$\displaystyle\int\limits^b_a f(x)\,dx$$ is the limit of this sum which is the exact area under the curve $$y=f(x)$$ from $$a$$ to $$b$$.  • If $$'f'$$ takes both positive and negative values then Riemann sum is sum of areas of rectangle that lie above $$x$$ axis and negative of area of rectangles that lie below $$x$$ axis.

Riemann sum = shaded area  $$(A_1)$$– unshaded area $$(A_2)$$

In terms of definite integral,

$$\displaystyle\int\limits^b_a f(x)\,dx=A_1–A_2$$  #### Let $$f(x)=x^2–3x$$, $$0\leq x\leq 4$$, find the Riemann sum with $$n=4$$ taking sample points to be right end points.

A 17

B 0

C 9

D –8

×

Riemann Sum (R.S) = $$\sum\limits^4_{i=1} f(x_i^*)\,\Delta x$$

$$\Delta x=\dfrac{4-0}{4}=\dfrac{4}{4}=1$$      $$\;x_i^*=x_i\to$$  right end points

$$\therefore\,R.S=f(x_1)\Delta x+f(x_2)\Delta x+f(x_3)\Delta x+f(x_4)\Delta x$$

$$=f(1)×1+f(2)×1+f(3)×1+f(4)×1$$

$$=(1–3)+(4–6)+(9–9)+(16–12)$$

$$=–2–2+0+4=0$$

This means that area above $$x$$ - axis is same as area below $$x$$ - axis. ### Let $$f(x)=x^2–3x$$, $$0\leq x\leq 4$$, find the Riemann sum with $$n=4$$ taking sample points to be right end points.

A

17

.

B

0

C

9

D

–8

Option B is Correct

# Evaluation of some Standard Sequences

The sigma notation is often used to conveniently write the sequences and series.

Consider

$$\sum\limits^n_{i=1}a_i=a_1+a_2+a_3+.......a_n$$

i.e.  write the term with values of variable of summative and then insert + sign.

(1)  $$\sum\limits^n_{i=1}c=cn$$ (where c is a constant independent of $$n$$)

(2)  $$\sum\limits^n_{i=1}ca_i=c\sum\limits^n_{i=1}a_i$$

(3)  $$\sum\limits^n_{i=1}a_i+b_i=\sum\limits^n_{i=1}a_i+\sum\limits^n_{i=1}b_i$$

(4)  $$\sum\limits^n_{i=1}a_i-b_i=\sum\limits^n_{i=1}a_i-\sum\limits^n_{i=1}b_i$$

(1)  $$\sum\limits^n_{i=1}\,i=1+2+3+4+.....n=\dfrac{n(n+1)}{2}$$

(Sum of first $$n$$ natural numbers is $$\dfrac{n(n+1)}{2}$$)

(2)  $$\sum\limits^n_{i=1}\,i^2=1^2+2^2+3^2+.....n^2=\dfrac{n(n+1)(2n+1)}{6}$$

(Sum of squares of first $$n$$ natural numbers is $$\dfrac{n(n+1)(2n+1)}{6}$$)

(3)  $$\sum\limits^n_{i=1}\,i^3=1^3+2^3+3^3+.....n^3=\left(\dfrac{n(n+1)}{2}\right)^2$$

Sum of cubes of first $$n$$ natural numbers is $$\left(\dfrac{n(n+1)}{2}\right)^2$$.

#### Find the value of the sum $$S=1^2+2^2+.....15^2$$ .

A 1240

B –89

C 700

D 26

×

$$1^2+2^2+......15^2=\sum\limits^{15}_{i=1}\,i^2$$

$$=\sum\limits^{15}_{i=1}\,i^2=\dfrac{15×16×31}{6}$$  $$\bigg[$$ Use $$\dfrac{n(n+1)(2n+1)}{6}$$ with $$n=15$$ $$\bigg]$$

$$=1240$$

### Find the value of the sum $$S=1^2+2^2+.....15^2$$ .

A

1240

.

B

–89

C

700

D

26

Option A is Correct