मराठी

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

प्रश्न

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 ______.

पर्याय

  • `"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
रिकाम्या जागा भरा

उत्तर

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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Integrals - Solved Examples [पृष्ठ १६०]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
पाठ 7 Integrals
Solved Examples | Q 24 | पृष्ठ १६०

व्हिडिओ ट्यूटोरियलVIEW ALL [3]

संबंधित प्रश्‍न

Evaluate `int_1^3(e^(2-3x)+x^2+1)dx`  as a limit of sum.


Evaluate the following definite integrals as limit of sums `int_(-1)^1 e^x dx`


Evaluate the following definite integrals as limit of sums.

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


Evaluate the definite integral:

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


Prove the following:

`int_0^1 xe^x dx = 1`


Prove the following:

`int_0^(pi/4) 2 tan^3 xdx = 1 - log 2`


Choose the correct answers The value of `int_0^1 tan^(-1)  (2x -1)/(1+x - x^2)` dx is 

(A) 1

(B) 0

(C) –1

(D) `pi/4`


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


` ∫  log x / x  dx `
 
 
 

\[\int\frac{4x + 3}{\sqrt{2 x^2 + 3x + 1}} dx\]

\[\int\frac{1 + \cos x}{\left( x + \sin x \right)^3} dx\]

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

 


\[\int\cot x \cdot \log \text{sin x dx}\]

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

\[\text{ ∫  cosec x  log}      \left( \text{cosec x} - \cot x \right) dx\]

\[\int x^3 \sin \left( x^4 + 1 \right) dx\]

\[\int \sec^4    \text{ x   tan x dx} \]

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

\[\int4 x^3 \sqrt{5 - x^2} dx\]

Evaluate the following integrals as limit of sums:

\[\int_1^3 \left( 3 x^2 + 1 \right)dx\]

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


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


Evaluate:

`int (sin"x"+cos"x")/(sqrt(9+16sin2"x")) "dx"`


Evaluate the following as limit of sum:

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


If f" = C, C ≠ 0, where C is a constant, then the value of `lim_(x -> 0) (f(x) - 2f (2x) + 3f (3x))/x^2` is


`lim_(n rightarrow ∞)1/2^n [1/sqrt(1 - 1/2^n) + 1/sqrt(1 - 2/2^n) + 1/sqrt(1 - 3/2^n) + ...... + 1/sqrt(1 - (2^n - 1)/2^n)]` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×