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.

 

Illustration Questions

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 \(g(x) = \int g(x) dx\)

           \(= \int x^{-1} dx\)

          \(= \displaystyle \int \dfrac{1}{x} dx\)

         \(= log _e x+C\) 

Illustration Questions

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

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 \(f(x) =cosec(x)\)

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

 

Illustration Questions

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 \(f(x) = \int f(x) dx\)

                \(= \int e^x \,dx\)

                \(= e^x + C\)

 

Illustration Questions

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

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

Illustration Questions

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

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.

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

 

Illustration Questions

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

image

This area can be verified  by 

Total area = Area of triangle + Area of rectangle 

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

\(= 4+8 = 12\, m^2\)

image image

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

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

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

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

 

Illustration Questions

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

Illustration Questions

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

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

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

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

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

image

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

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