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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.
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