0) then the value of the integral ππgf∫-π4π4g(f(x))dx is ______. - | Shaalaa.com" /> 0) then the value of the integral ππgf∫-π4π4g(f(x))dx is ______. " /> 0) then the value of the integral ππgf∫-π4π4g(f(x))dx is ______., Properties of Definite Integrals" />
हिंदी

If f(x) = 2-xcosx2+xcosx and g(x) = log"e"x, (x > 0) then the value of the integral ππgf∫-π4π4g(f(x))dx is ______. -

Advertisements
Advertisements

प्रश्न

If f(x) = `(2 - xcosx)/(2 + xcosx)` and g(x) = logex, (x > 0) then the value of the integral `int_((-π)/4)^(π/4) "g"("f"(x))"d"x` is ______.

विकल्प

  • loge1

  • loge2

  • logee

  • loge3

MCQ
रिक्त स्थान भरें

उत्तर

If f(x) = `(2 - xcosx)/(2 + xcosx)` and g(x) = logex, (x > 0) then the value of the integral `int_((-π)/4)^(π/4) "g"("f"(x))"d"x` is `underlinebb(log_e1)`.

Explanation:

f(x) = `(2 - xcosx)/(2 + xcosx)`       `{:("g"(x) = log_"e"x),("h"(x > 0)):}`

`int_((-π)/4)^(π/4) "g"("f"(x))"d"x`

`int_((-π)/4)^(π/4) log_"e"((2 - xcosx)/(2 + xcosx))"d"x`

Let h(x) = `log_"e"  (2 - xcosx)/(2 + x cosx)`

h(–x) = `log_"e"  (2 + xcosx)/(2 - x cosx)`

= `log_"e"((2 - xcosx)/(2 + xcosx))^-1`

= `-log_"e"  (2 - xcosx)/(2 + xcosx)`  = –f(x)

Odd function

For odd function

`int_((-π)/4)^(π/4)"h"(x)` = 0 = loge1

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×