English

If f and g are continuous functions in [0, 1] satisfying f(x) = f(a – x) and g(x) + g(a – x) = a, then afgd∫0af(x)⋅g(x)dx is equal to ______. - Mathematics

Advertisements
Advertisements

Question

If f and g are continuous functions in [0, 1] satisfying f(x) = f(a – x) and g(x) + g(a – x) = a, then `int_0^"a" "f"(x) * "g"(x)"d"x` is equal to ______.

Options

  • `"a"/2`

  • `"a"/2 int_0^"a" "f"(x)"d"x`

  • `int_0^"a" "f"(x)"d"x`

  • `"a" int_0^"a" "f"(x)"d"x`

MCQ
Fill in the Blanks

Solution

If f and g are continuous functions in [0, 1] satisfying f(x) = f(a – x) and g(x) + g(a – x) = a, then `int_0^"a" "f"(x) * "g"(x)"d"x` is equal to `"a"/2 int_0^"a" "f"(x)"d"x`.

Explanation:

Since I = `int_0^"a" "f"(x) * "g"(x)"d"x`

= `int_0^"a" "f"("a" - x) "g"("a" - x)"d"x`

= `int_0^"a" "f"(x)("a" - "g"(x))"d"x`

= `"a" int_0^"a" "f"(x) "d"x - int_0^"a" "f"(x) * "g"(x)"d"x`

= `"a" int_0^"a" "f"(x)"d"x - 1`

or 1 = `"a"/2 int_0^"a" "f"(x)"d"x`

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Integrals - Solved Examples [Page 160]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 12
Chapter 7 Integrals
Solved Examples | Q 24 | Page 160

RELATED QUESTIONS

Evaluate `int_(-1)^2(e^3x+7x-5)dx` as a limit of sums


Evaluate the following definite integrals as limit of sums.

`int_0^4 (x + e^(2x)) dx`


Evaluate the definite integral:

`int_(pi/2)^pi e^x ((1-sinx)/(1-cos x)) dx`


Evaluate the definite integral:

`int_0^(pi/4) (sinx cos x)/(cos^4 x + sin^4 x)`dx


Evaluate the definite integral:

`int_0^(pi/2) (cos^2 x dx)/(cos^2 x + 4 sin^2 x)`


Prove the following:

`int_1^3 dx/(x^2(x +1)) = 2/3 + log  2/3`


Prove the following:

`int_0^1 xe^x dx = 1`


Evaluate  `int_0^1 e^(2-3x) dx` as a limit of a sum.


if `int_0^k 1/(2+ 8x^2) dx = pi/16` then the value of k is ________.

(A) `1/2`

(B) `1/3`

(C) `1/4`

(D) `1/5`


Evaluate : `int_1^3 (x^2 + 3x + e^x) dx` as the limit of the sum.


\[\int\frac{\sin^3 x}{\sqrt{\cos x}} dx\]

\[\int\frac{1}{x} \left( \log x \right)^2 dx\]


\[\int\frac{\log x^2}{x} dx\]

\[\int\sec x \cdot \text{log} \left( \sec x + \tan x \right) dx\]

\[\int\frac{1}{x\sqrt{x^4 - 1}} dx\]

\[\int\limits_0^\pi \frac{\sin x}{\sin x + \cos x} dx\]

\[\int\frac{\sqrt{\tan x}}{\sin x \cos x} dx\]


Evaluate `int_1^4 ( 1+ x +e^(2x)) dx` as limit of sums.


Evaluate the following as limit of sum:

`int_0^2 "e"^x "d"x`


Evaluate the following:

`int_0^2 ("d"x)/("e"^x + "e"^-x)`


Evaluate the following:

`int_0^(pi/2) (tan x)/(1 + "m"^2 tan^2x) "d"x`


Evaluate the following:

`int_0^pi x sin x cos^2x "d"x`


Evaluate the following:

`int_(pi/3)^(pi/2) sqrt(1 + cosx)/(1 - cos x)^(5/2)  "d"x`


The value of `lim_(x -> 0) [(d/(dx) int_0^(x^2) sec^2 xdx),(d/(dx) (x sin x))]` is equal to


The limit of the function defined by `f(x) = {{:(|x|/x",", if x ≠ 0),(0",", "otherwisw"):}`


`lim_(x -> 0) (xroot(3)(z^2 - (z - x)^2))/(root(3)(8xz - 4x^2) + root(3)(8xz))^4` is equal to


Let f: (0,2)→R be defined as f(x) = `log_2(1 + tan((πx)/4))`. Then, `lim_(n→∞) 2/n(f(1/n) + f(2/n) + ... + f(1))` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×