Find the best tutors and institutes for Class 12 Tuition
Search in
By using the properties of definite integrals, evaluate the integrals

Adding (1) and (2), we obtain

By using the properties of definite integrals, evaluate the integrals

Adding (1) and (2), we obtain

By using the properties of definite integrals, evaluate the integrals

Adding (1) and (2), we obtain

By using the properties of definite integrals, evaluate the integrals

Adding (1) and (2), we obtain

By using the properties of definite integrals, evaluate the integrals
![]()
It can be seen that (x + 2) ≤ 0 on [−5, −2] and (x + 2) ≥ 0 on [−2, 5].

By using the properties of definite integrals, evaluate the integrals
![]()
It can be seen that (x − 5) ≤ 0 on [2, 5] and (x − 5) ≥ 0 on [5, 8].

By using the properties of definite integrals, evaluate the integrals

By using the properties of definite integrals, evaluate the integrals

By using the properties of definite integrals, evaluate the integrals

By using the properties of definite integrals, evaluate the integrals

Adding (1) and (2), we obtain

By using the properties of definite integrals, evaluate the integrals
![]()
As sin2 (−x) = (sin (−x))2 = (−sin x)2 = sin2x, therefore, sin2x is an even function.
It is known that if f(x) is an even function, then ![]()

By using the properties of definite integrals, evaluate the integrals
![]()

Adding (1) and (2), we obtain

By using the properties of definite integrals, evaluate the integrals
![]()
As sin7 (−x) = (sin (−x))7 = (−sin x)7 = −sin7x, therefore, sin2x is an odd function.
It is known that, if f(x) is an odd function, then ![]()
![]()
By using the properties of definite integrals, evaluate the integrals

It is known that,

![]()
![]()
By using the properties of definite integrals, evaluate the integrals
![]()

Adding (1) and (2), we obtain

By using the properties of definite integrals, evaluate the integrals
![]()

Adding (1) and (2), we obtain

sin (π − x) = sin x
![]()
![]()
Adding (4) and (5), we obtain

Let 2x = t ⇒ 2dx = dt
When x = 0, t = 0
and when x=π2, t=πx=π2, t=π
∴ I=12∫π0log sin tdt−π2log 2I=12∫0πlog sin tdt-π2log 2
⇒I=I2−π2log 2 [from 3]⇒I=I2-π2log 2 [from 3]
⇒I2=−π2log 2⇒I2=-π2log 2
⇒I=−πlog 2
By using the properties of definite integrals, evaluate the integrals
![]()
It is known that, ![]()
![]()
Adding (1) and (2), we obtain

By using the properties of definite integrals, evaluate the integrals
![]()
It can be seen that, (x − 1) ≤ 0 when 0 ≤ x ≤ 1 and (x − 1) ≥ 0 when 1 ≤ x ≤ 4

Show that
if f and g are defined as
and
![]()

Adding (1) and (2), we obtain

The value of
is
A. 0
B. 2
C. π

It is known that if f(x) is an even function, then
and
if f(x) is an odd function, then ![]()

Hence, the correct answer is C.
The value of
is
A. 2
B. ![]()
C. 0
D. ![]()
![]()

Adding (1) and (2), we obtain

Hence, the correct answer is C.
How helpful was it?
How can we Improve it?
Please tell us how it changed your life *
Please enter your feedback
UrbanPro.com helps you to connect with the best Class 12 Tuition in India. Post Your Requirement today and get connected.
Find best tutors for Class 12 Tuition Classes by posting a requirement.
Get started now, by booking a Free Demo Class