हिंदी

If F (A + B − X) = F (X), Then B ∫ a X F (X) Dx is Equal To,A + B 2 B ∫ a F ( B − X ) D X,A + B 2 B ∫ a F ( B + X ) D X,B − a 2 B ∫ a F ( X ) D X,B + a 2 B ∫ a F ( X ) D X - Mathematics

Advertisements
Advertisements

प्रश्न

If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to

विकल्प

  • \[\frac{a + b}{2} \int\limits_a^b f\left( b - x \right) dx\]

     

  • \[\frac{a + b}{2} \int\limits_a^b f\left( b + x \right) dx\]

     

  • \[\frac{b - a}{2} \int\limits_a^b f\left( x \right) dx\]
  • \[\frac{b + a}{2} \int\limits_a^b f\left( x \right) dx\]
MCQ

उत्तर

\[\frac{a + b}{2} \int\limits_a^b f\left( x \right) dx\]

\[Let\, I = \int_a^b x f\left( x \right) d x .............(1)\]

\[ = \int_a^b \left( a + b - x \right) f\left( a + b - x \right) d x\]

\[ = \int_a^b \left( a + b - x \right) f\left( x \right) dx ...............(2)\]

\[ \text{Adding (1) and (2)}\]

\[2I = \int_a^b \left( x + a + b - x \right) f\left( x \right) d x\]

\[ = \left( a + b \right) \int_a^b f\left( x \right) d x \]

\[Hence\ I = \frac{a + b}{2} \int_a^b f\left( x \right) d x\]

shaalaa.com
Definite Integrals
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 20: Definite Integrals - MCQ [पृष्ठ १२०]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
MCQ | Q 39 | पृष्ठ १२०

संबंधित प्रश्न

\[\int\limits_2^3 \frac{x}{x^2 + 1} dx\]

\[\int\limits_0^{\pi/2} \left( a^2 \cos^2 x + b^2 \sin^2 x \right) dx\]

\[\int\limits_0^{\pi/2} x^2 \cos\ 2x\ dx\]

\[\int\limits_0^{\pi/2} x^2 \cos^2 x\ dx\]

\[\int\limits_0^1 \left( x e^{2x} + \sin\frac{\ pix}{2} \right) dx\]

\[\int\limits_0^{2\pi} e^x \cos\left( \frac{\pi}{4} + \frac{x}{2} \right) dx\]

\[\int_0^\pi e^{2x} \cdot \sin\left( \frac{\pi}{4} + x \right) dx\]

\[\int_0^\frac{\pi}{4} \left( a^2 \cos^2 x + b^2 \sin^2 x \right)dx\]

\[\int\limits_4^{12} x \left( x - 4 \right)^{1/3} dx\]

\[\int\limits_0^{\pi/2} x^2 \sin\ x\ dx\]

\[\int\limits_1^2 \frac{1}{x \left( 1 + \log x \right)^2} dx\]

\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{3/2}} dx\]

\[\int_\frac{1}{3}^1 \frac{\left( x - x^3 \right)^\frac{1}{3}}{x^4}dx\]

\[\int_0^2 2x\left[ x \right]dx\]

\[\int\limits_0^{\pi/2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} dx\]

\[\int\limits_0^a \frac{1}{x + \sqrt{a^2 - x^2}} dx\]

\[\int\limits_0^\pi x \log \sin x\ dx\]

\[\int\limits_3^5 \left( 2 - x \right) dx\]

\[\int\limits_1^2 \left( x^2 - 1 \right) dx\]

\[\int\limits_0^{\pi/2} \cos x\ dx\]

\[\int\limits_0^2 \left( x^2 + 2x + 1 \right) dx\]

\[\int\limits_2^3 x^2 dx\]

\[\int\limits_0^1 \frac{1}{x^2 + 1} dx\]

\[\int\limits_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} dx, n \in N .\]

\[\int\limits_0^1 \frac{2x}{1 + x^2} dx\]

Evaluate each of the following  integral:

\[\int_0^1 x e^{x^2} dx\]

 


Evaluate each of the following integral:

\[\int_e^{e^2} \frac{1}{x\log x}dx\]

The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is 


The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]

 


The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is

 


If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\]  then the value of I10 + 90I8 is

 


The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .


\[\int\limits_0^{15} \left[ x^2 \right] dx\]


\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]


Evaluate the following:

`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`


Evaluate the following:

`int_(-1)^1 "f"(x)  "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x  < 0):}`


Choose the correct alternative:

Γ(n) is


Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1


`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×