Informative line

### Preview Of Integral Calculus

Learn meaning of antiderivative and graphical meaning of integral calculus formulas. Practice problems for calculating antiderivative of x^n, trigonometric, exponential and average value of function.

# Meaning of Antiderivative

• A function $$F(x)$$ is an antiderivative of the function $$f(x)$$,

if  $$\dfrac{d}{dx} F(x) = f(x)$$

• To indicate $$F(x)$$ as an antiderivative of $$f(x)$$, the following notation is  used

$$F(x) = \int f(x) dx$$

• In General :

As  $$\dfrac{d}{dx} [F(x) +C]\;=f(x)$$

so, $$\int f(x) dx= F(x) +C$$

where C is a constant of integration.

• Antiderivatives are also known as indefinite integrals.

#### If a function $$g(x)$$ is a derivative of function $$f(x)$$, the value of $$\int g(x) dx$$ will be

A $$g^2(x)$$

B $$g(x)$$

C $$f(x)$$

D $$\dfrac{f(x)}{2}$$

×

As $$\dfrac{d}{dx} f(x) = g(x)$$

$$\therefore$$ By definition of antiderivative

$$\int g(x) dx = f(x)$$

Hence, option (C) is correct.

### If a function $$g(x)$$ is a derivative of function $$f(x)$$, the value of $$\int g(x) dx$$ will be

A

$$g^2(x)$$

.

B

$$g(x)$$

C

$$f(x)$$

D

$$\dfrac{f(x)}{2}$$

Option C is Correct

# Antiderivative of xn

Case 1 :

• Antiderivative of $$x^n$$ when $$n \neq -1$$
• Let $$f(x) = x^n$$
• Antiderivative of $$f(x)= \int f(x) dx$$

= $$\int x^n dx$$

$$= \dfrac{x^{n+1}}{n+1} +C$$    where $$n \neq -1$$

Case 2 :

• Antiderivative of $$x^n$$ when n= 1
• Let $$g(x)=x^{-1}$$
• Antiderivative of $$g(x) = \int g(x) dx$$

$$= \int x^{-1} dx$$

$$= \displaystyle \int \dfrac{1}{x} dx$$

$$= log _e x+C$$

#### Which pair of function and its antiderivative is incorrect?

A $$\displaystyle \int x^5 dx = \dfrac{x^6}{6} + C$$

B $$\displaystyle \int x^{-1} dx = log_e x + C$$

C $$x^{-3} dx = \dfrac{-1}{2x^2} + C$$

D $$\displaystyle \int x^{10} dx = 10\,log_e x+ C$$

×

As $$\displaystyle \int x^n dx = \dfrac{x^{n+1}}{n+1} + C$$

$$\therefore\displaystyle \int x^5 dx = \dfrac{x^{5+1}}{5+1} + C$$

$$=\dfrac{x^6}{6} +C$$

Hence, option (A) is correct.

$$\therefore\displaystyle \int x^{-1} dx = \displaystyle \int\dfrac{1}{x} dx$$

$$= log _e x+C$$

Hence, option (B) is correct.

$$\therefore\displaystyle \int x^{-3} dx = \dfrac{x^{-3+1}}{-3+1}$$

$$= \dfrac{x^{-2}}{-2} = \dfrac{-1}{2x^2} + C$$

Hence, option (C) is correct.

$$\therefore\displaystyle \int x^{10} dx = \dfrac{x^{10+1}}{10+1} +C$$

$$= \dfrac{x^{11}}{11} +C$$

Hence, option (D) is incorrect.

### Which pair of function and its antiderivative is incorrect?

A

$$\displaystyle \int x^5 dx = \dfrac{x^6}{6} + C$$

.

B

$$\displaystyle \int x^{-1} dx = log_e x + C$$

C

$$x^{-3} dx = \dfrac{-1}{2x^2} + C$$

D

$$\displaystyle \int x^{10} dx = 10\,log_e x+ C$$

Option D is Correct

# Definite Integration

• Since, velocity is the derivative of displacement so, displacement is the antiderivative of velocity $$v$$.
• Considering velocity as a function of time and taking very small interval of time, then the summation of product of these small intervals with velocity gives the displacement.
• For the calculation of displacement between time interval t1 to t2, the summation is denoted as

$$\displaystyle\int\limits_{t_1}^{t_2} v . dt =$$Displacement

• In general

Let   $$\dfrac{d}{dx} F(x) = f(x)$$

then, antiderivative  $$\int f(x)\, dx = F(x) +C$$

• Performing this integration for the limits a to b

$$\displaystyle\int\limits_{a}^{b} f(x) dx = [F(x) + C]^b_a$$

$$= [F(b) + C] - [F(a) + C]$$

$$\displaystyle\int\limits_{a}^{b} f(x)dx = [F(b)-F(a) ]$$

Conclusion : $$\displaystyle\int\limits_{a}^{b} f(x) dx$$ represents summation of small units into the end points [a,b],

where a is the lower limit

b is the upper limit

#### If velocity of particle is given by function of time v(t) = 3t2+2. Find the displacement of particle over the period 0 to 4 sec.

A 24 m

B 64 m

C 53 m

D 72 m

×

Since, velocity is the derivative of displacement

So, displacement is the antiderivative of velocity$$v$$.

$$\displaystyle\int\limits_{t_1}^{t_2} v . dt =$$ Displacement

Given : v(t) = 3t2+2, t1 = 0, t2 = 4 sec

Displacement $$=\displaystyle\int\limits_{0}^{4}(3t^2 + 2)\, dt$$

$$= \left[\dfrac{3t^3}{3} + 2t\right]_0^4$$

= $$[t^3 + 2t]^4_0$$

$$= [(4)^3 + 2× 4] - [0^3 + 2× 0]$$

$$= 72 \,m$$

### If velocity of particle is given by function of time v(t) = 3t2+2. Find the displacement of particle over the period 0 to 4 sec.

A

24 m

.

B

64 m

C

53 m

D

72 m

Option D is Correct

# Average Value of Function f(x) Over the Given Interval

• Consider a function $$y= f(x)$$, as shown in figure.

• The area under the curve is given by the definition of definite integral.

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

• Average value of the function can be calculated as

Area of PQRS $$\cong$$  Area under the curve

$$f_{avg} × (b-a) = \displaystyle\int\limits ^b_a f(x) dx$$

$$f_{avg} = \dfrac{\displaystyle \int \limits^b_af(x)dx}{(b-a)}$$

#### Find the average value of function $$y= 3x^2$$ over the interval $$2\leq x< 4$$.

A 36

B 28

C 42

D 18

×

The average value of function $$y= f(x)$$ over the internal a to b is given as

$$\overline f= \dfrac{1}{(b-a)} \displaystyle\int\limits_a^b f(x)dx$$

Given : $$f(x) = 3x^2$$, a =2, b =4

$$\overline f= \dfrac{1}{(4-2)} \displaystyle\int\limits_2^4 3x^2\,dx$$

$$\overline f= \dfrac{1}{2} \left[x^3\right]_2^4$$

$$\overline f= \dfrac{1}{2} \left[4^3 -2^3\right]$$

$$\overline f= \dfrac{1}{2} \left[64-8\right]$$

$$\overline f= \dfrac{1}{2} × 56 = 28$$

### Find the average value of function $$y= 3x^2$$ over the interval $$2\leq x< 4$$.

A

36

.

B

28

C

42

D

18

Option B is Correct

# Antiderivative  of  Trigonometric Functions

(1)  If $$f(x) = sin(x)$$

$$\int f(x) \,dx = \int sin x \,dx = - cos \,x +C$$

(2)   If  $$f(x) = cos(x)$$

$$\int f(x)\, dx = \int cos \,x \,dx = sin \,x +C$$

(3)  If $$f(x) =tan(x)$$

$$\int f(x)\, dx = \int tan \,x \,dx = \ell n |sec \,x| +C$$

(4)  If $$f(x) =cot(x)$$

$$\int f(x) \,dx = \int cot\,x \,dx = \ell n |sin \,x| +C$$

(5)  If $$f(x) =sec(x)$$

$$\int f(x) \,dx = \int sec \,x \,dx = \ell n |sec \,x + tan\,x| +C$$

(6) If $$f(x) =cosec(x)$$

$$\int f(x) \,dx = \int cosec \,x \,dx = \ell n |cosec \,x + cot \,x| +C$$

#### Which pair of function and its antiderivative is incorrect?

A $$\displaystyle \int tan\,5x \,dx = \dfrac{1}{5} \ell n |sec \,5x| +C$$

B $$\displaystyle \int cosec\,2x \,dx = \dfrac{1}{2} \ell n |cosec\,2x \,-cot\,2x| +C$$

C $$\displaystyle \int cos\,3x \,dx = \dfrac{1}{3} |sin \,3x| +C$$

D $$\displaystyle \int sec\,3x \,dx = 3\, |-cosec \,x + tan \,x| +C$$

×

For $$\int tan \, 5x \,dx$$

Let $$t = 5x$$

$$dt = 5\,dx$$

$$dx = \dfrac{dt}{5}$$

So, $$\displaystyle \int tan \,t × \dfrac{dt}{5}$$

$$=\displaystyle\dfrac{1}{5} \int tan \,t \, dt$$

$$= \dfrac{1}{5} \ell n |sec \,t | + C$$

$$\left[\therefore\displaystyle\int tan \,x \, dx = \ell n |sec x| +C\right]$$

put $$t = 5x$$

$$= \dfrac{1}{5} \ell n |sec \,(5x) | + C$$

Hence, option (A) is correct.

For $$\displaystyle \int cosec \,2x \,dx$$

Let $$t = 2x$$

$$dt = 2dx$$

$$dx = \dfrac{dt}{2}$$

So, $$\displaystyle \int cosec \,t × \dfrac{dt}{2}$$

=$$\dfrac{1}{2}\displaystyle \int cosec \,t \, dt$$

$$\left[\therefore\displaystyle\int cosec \, x \, dx = \ell n |cosec \,x-cot \,x| +C\right]$$

$$= \dfrac{1}{2} \ell n |cosec \,t-cot \,t| +C$$

put $$t =2x$$

$$= \dfrac{1}{2} \ell n |cosec \,2x-cot \,2x| +C$$

Hence, option (B) is correct.

For $$\int cos \,3x \,dx$$

Let $$t= 3x$$

$$dt = 3\,dx$$

$$dx = \dfrac{dt}{3}$$

So, $$\displaystyle \int cos \,t .\dfrac{dt}{3}$$

$$\dfrac{1}{3}\displaystyle \int cos \,t\; dt$$

$$\dfrac{1}{3} sin \,t+C$$

$$\left[\therefore\displaystyle\int cos \,t \, dt = \,sin\,t +C\right]$$

put $$t= 3x$$

$$= \dfrac{1}{3} sin \,3x +C$$

Hence, option (C) is correct.

For $$\int sec \,3x \,dx$$

Let $$t= 3x$$

$$dt = 3\,dx$$

$$dx = \dfrac{dt}{3}$$

So, $$\displaystyle \int sec \,t .\dfrac{dt}{3}$$

$$\dfrac{1}{3}\displaystyle \int sec \,t \;dt$$

$$\dfrac{1}{3} \ell n |sec\,t + tan \,t|\,+C$$

$$\left[\therefore\displaystyle\int sec \,x \, dx = \ell n|\,sec \,x +tan\,x| +C\right]$$

put $$t= 3x$$

$$\dfrac{1}{3} \ell n |sec\,3x + tan \,3x|\,+C$$

Hence, option (D) is incorrect.

### Which pair of function and its antiderivative is incorrect?

A

$$\displaystyle \int tan\,5x \,dx = \dfrac{1}{5} \ell n |sec \,5x| +C$$

.

B

$$\displaystyle \int cosec\,2x \,dx = \dfrac{1}{2} \ell n |cosec\,2x \,-cot\,2x| +C$$

C

$$\displaystyle \int cos\,3x \,dx = \dfrac{1}{3} |sin \,3x| +C$$

D

$$\displaystyle \int sec\,3x \,dx = 3\, |-cosec \,x + tan \,x| +C$$

Option D is Correct

# Antiderivative of Exponential Function

• Consider an exponential function $$f(x) = e^x$$
•  Antiderivative of $$f(x) = \int f(x) dx$$

$$= \int e^x \,dx$$

$$= e^x + C$$

#### What will be the antiderivative of function $$f(x) = e^{2x}$$?

A $$e^x +C$$

B $$e^{2x} +C$$

C $$e^{4x} +C$$

D $$\dfrac{1}{2}e^{2x} +C$$

×

Given : $$f(x) = e^{2x}$$

Let $$2x = t$$

$$dt = 2\,dx$$

$$dx = \dfrac{dt}{2}$$

Antiderivative of $$f(x) = \int f(x) dx$$

$$= \int e^{2x} dx$$

$$= \displaystyle \int e^t \, \dfrac{dt}{2}$$

$$= \dfrac{1}{2}\displaystyle \int e^t \, dt$$

$$=\dfrac{1}{2} e^t + C$$

$$[\therefore \int e^x \,dx = e^x]$$

put  $$t= 2x$$

$$= \dfrac{1}{2} e^{2x} +C$$

Hence, option (D) is correct.

### What will be the antiderivative of function $$f(x) = e^{2x}$$?

A

$$e^x +C$$

.

B

$$e^{2x} +C$$

C

$$e^{4x} +C$$

D

$$\dfrac{1}{2}e^{2x} +C$$

Option D is Correct

# Graphical Meaning of Integration

• Consider a continuous function  $$y = f(x)$$ defined between limits $$x=a$$ to $$x =b$$, as shown in figure with a graph of any arbitrary shape.
• The graph is bounded by curve $$y = f(x)$$$$y= 0$$ and the lines  $$x=0$$ and $$x=b$$.

• Consider a strip of thickness $$dx$$ at a distance $$x$$ from the origin with length  parallel to $$y{-axis}$$, as shown in figure.
• Area of the shaded region is

$$dA = f (x)\, dx$$

• If the area is divided into 'n' strip of equal width, then complete area of the region below the graph and above the $$x$$ - axis confined between the coordinates at $$x=a$$ to $$x=b$$ can be obtained by summing up the area of 'n' individual strip.
• For better approximation, 'n' is considered to be very large.
• This summation is represented by definite integral.

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

#### Applying the method of integration, find the area between the line $$y=2x$$, $$x-axis$$ and the ordinates at $$x=$$2 m to $$x=$$ 4 m.

A 7 m2

B 10 m2

C 12 m2

D 8 m2

×

Area under the line

$$A = \displaystyle \int\limits _2^4 f(x) dx$$

$$A = \displaystyle \int\limits _2^4 2x\, dx$$

$$A = \left[\dfrac{2x^2}{2} \right ]^4_2$$

A= [42 – (2)2 ] = 12 m2

This area can be verified  by

Total area = Area of triangle + Area of rectangle

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

$$= 4+8 = 12\, m^2$$

### Applying the method of integration, find the area between the line $$y=2x$$, $$x-axis$$ and the ordinates at $$x=$$2 m to $$x=$$ 4 m.

A

7 m2

.

B

10 m2

C

12 m2

D

8 m2

Option C is Correct

#### The force applied on a body is given by a function $$F =2x$$. Calculate the work done to displace the body from 0 to 5 m. [Work done = Force × displacement]

A 15 J

B 30 J

C 50 J

D 25 J

×

Work done = Area under the graph

$$W = \displaystyle\int\limits ^5_0 F.dx$$

$$W = \displaystyle\int\limits ^5_0 2x.dx$$

$$W = 2\left[\dfrac{x^2}{2}\right]^5_0$$

$$W = [5^2 - 0^2] = 25 \,J$$

### The force applied on a body is given by a function $$F =2x$$. Calculate the work done to displace the body from 0 to 5 m. [Work done = Force × displacement]

A

15 J

.

B

30 J

C

50 J

D

25 J

Option D is Correct