PRACTICE QUESTIONS ON PROPERTIES OF DEFINITE INTEGRALS

Evaluate the following problems using properties of integration. 

Question 1 :

Solution :

Let f(x) = sin x cos⁴ x

To check if the function is odd or even, we use the rule

Put x = -x

f(-x) = sin (-x) [cos (-x)]⁴

= -sin cos⁴ x

f(-x) = -f(x)

Since the function is odd, the value of the integration will become 0.

Question 2 :

Solution :

Let f(x) = x³ cos³x

By observing both upper limit and lowe limit, we know that both are the same. Now we have to check whether the given function is odd or even.

Put x = -x

f(-x) = (-x)³ [cos(-x)]³

f(-x) = -x³ [cos³ x]

f(-x) = -f(x)

Since it is odd function, the integrating will become 0. 

Question 3 :

Solution :

Since the upper and lower limit are not the same, we are going to check if the function is odd or even. 

Adding (1) and (2), we get

Multiplying and dividing by 2, we get

Integrating and applying the limits, we get

2 I = (1/2) [-cos 2x/2] 0 to π/2.

2 I = (1/4) [-cos 2x] 0 to π/2

Upper limit = π/2 and lower limit = 0

2 I = (1/4)[-cos 2(π/2) + cos 0]

= (1/4)[-cos π + cos 0]

= (1/4)[-(-1) + 1]

= (1/4)[2]

2I = 1/2

I = 1/4

So, the value of the given integration is 1/4.𝜋Upeer limit = 

Question 4 :

Solution ;

Let f(x) = cos³x

Put x = -x

f(-x) = [cos (-x)]³

= cos³ x

f(-x) = f(x)

Since it is even function, 

Using the formula in trigonometry,

= 1/2[-1/3 + 3(1)]

= 1/2[-1/3 + 3]

= 1/2[(-1 + 9)/3]

= 1/2[8/3]

= 4/3

Question 5 :

Solution :

Let f(x) = sin²x cos x

Put x = -x

f(-x) = [sin (-x)]² cos (-x)

= [-sin x]² cos x

= sin² x cos x

f(-x) = f(x)

So, the given function is even.

The function f(x) is even.

cos 2x cos x = (1/2)[2cos 2x cos x]

= (1/2)[cos (2x + x) + cos (2x - x)]

= (1/2)[cos 3x + cos x]

Integrating and applying the limit, we get

= [(1/2) sin x - (1/2) (sin 3x/3)] 0 to π/2

Upper limit = π/2 and lower limit = 0

= (1/2) (sin π/2 - sin 0) - (1/6)(sin 3π/2)

= (1/2) (1) - (1/6)(-1)

= (1/2) + (1/6)

= (3 + 1)/6

= 4/6

= 2/3

Question 6 :

Solution :

Let f(x) = x sin² x

Put x = -x

f(-x) = -x [sin(-x)]²

= -x [-sin x]²

= -x sin² x

f(-x) = - f(x)

Since it is odd function, the integration will become 0. That is ,

Question 7 :

Solution :

Using the property of integration

Adding (1) and (2), we get

2I  =  0

I  =  0

So, the answer is 0.

Question 8 :

Using the property of integration, we get

Adding (1) and (2), we get

Intergrating and applying the limit, we get

= [x] 0 to 3

Upper limit = 3 and lower limit = 0

2I = 3 - 0

2I = 3

I = 3/2

Question 9 :

Let f(x) = x³ sin²x

Put x = -x

f(-x) = (-x)³ sin²(-x)

= (-x)³ [-sin x]²

= -x³ sin²x

f(-x) = -f(x)

Since it is odd function, the integration will become 

Question 10 :

Let f(x) = log [(3 - x)/(3 + x)]

= log (3 - x) - log (3 + x)

Put x = -x

f(-x) = log (3 - (-x)) - log (3 - x)

= log (3 + x) - log (3 - x)

= - [log (3 - x) + log (3 + x)]

f(-x) = -f(x)

Since it is odd function, 

Question 11 :

Solution :

Let f(x) = x sin x

Put x = -x

f(-x) = -x sin (-x)

= -x (-sin x)

= x sin x

f(-x) = f(x)

Since it is even function, using the property

u = x and dx = sin x

udv = uv - ∫ v du

du = dx and v = -cos x

= 2

Question 12 :

Solution :

Let f(x) = sin²x

Put x = -x

f(-x) = [sin(-x)]²

= [-sin x]²

= -sin² x

f(-x) = -f(x)

Since it is odd function, the value of the integration will be 

Question 13 :

Solution :

Adding (1) and (2), we get

Integrating dx, we get x. By applying the limits, we get 

= π/2 - 0

2I = π/2

I = π/4

Question 14 :

Solution :

Integrating and applying the limit, we get

= x^{(n + 1)}/(n + 1) - x^{(n + 2)}/(n + 2)

= x^{(n + 1)}/(n + 1) - x^{(n + 2)}/(n + 2)

Applying the upper limit = 1 and lower limit = 0

= 1/(n + 1) - 1/(n + 2)

= [(n + 2) - (n + 1)]/(n + 1)(n + 2)

= 1/(n + 1)(n + 2)

Question 15 :

Solution :

Adding (1) and (2), we get

The vcalue of log 1 is 0.

= 0

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.