हिंदी

If F is an Integrable Function, Show that (I) a ∫ − a F ( X 2 ) D X = 2 a ∫ 0 F ( X 2 ) D X - Mathematics

Advertisements
Advertisements

प्रश्न

If f is an integrable function, show that

\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]

योग
Advertisements

उत्तर

\[I = \int_{- a}^a f\left( x^2 \right) d x\]
\[Here\ g\left( x \right) = f( x^2 )\]
\[ \Rightarrow g\left( - x \right) = f \left( - x \right)^2 = f( x^2 ) = g\left( x \right) i.e, g\left( x \right) \text{is even} \]
Therefore
\[I = 2 \int_0^a f\left( x^2 \right) d x .............\left[\text{Using }\int_{- a}^a g\left( x \right) d x = 2 \int_0^a g\left( x \right) dx \text{ when }g\left( x \right) \text{is even} \right]\]

shaalaa.com
Definite Integrals
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 20: Definite Integrals - Exercise 20.5 [पृष्ठ ९६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Exercise 20.5 | Q 45.1 | पृष्ठ ९६

संबंधित प्रश्न

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

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

\[\int\limits_0^1 \sqrt{x \left( 1 - x \right)} dx\]

\[\int\limits_1^4 \frac{x^2 + x}{\sqrt{2x + 1}} dx\]

\[\int\limits_0^1 x \left( 1 - x \right)^5 dx\]

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

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

\[\int_0^1 \frac{1}{1 + 2x + 2 x^2 + 2 x^3 + x^4}dx\]

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

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

\[\int\limits_0^1 \frac{1 - x^2}{x^4 + x^2 + 1} dx\]

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

\[\int\limits_0^{\pi/2} \frac{\sin x \cos x}{\cos^2 x + 3 \cos x + 2} dx\]

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

\[\int_0^{2\pi} \cos^{- 1} \left( \cos x \right)dx\]

Evaluate the following integral:

\[\int_{- a}^a \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right)d\theta\]

\[\int\limits_1^3 \left( 3x - 2 \right) dx\]

\[\int\limits_1^2 x^2 dx\]

\[\int\limits_1^4 \left( 3 x^2 + 2x \right) dx\]

\[\int\limits_0^2 \left( 3 x^2 - 2 \right) dx\]

\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]

\[\int\limits_0^\infty e^{- x} dx .\]

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{2} e^x \left( \sin x - \cos x \right)dx\]

 


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

\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.  

 

\[\int\limits_0^1 2^{x - \left[ x \right]} dx\]

\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]


\[\int\limits_0^{\pi/4} e^x \sin x dx\]


\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]


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


Using second fundamental theorem, evaluate the following:

`int_0^1 "e"^(2x)  "d"x`


Using second fundamental theorem, evaluate the following:

`int_1^2 (x - 1)/x^2  "d"x`


Evaluate the following using properties of definite integral:

`int_0^1 x/((1 - x)^(3/4))  "d"x`


Evaluate the following:

`int_0^oo "e"^(- x/2) x^5  "d"x`


Choose the correct alternative:

If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x)  "d"x + int_"c"^"b" f(x)  "d"x` is


Choose the correct alternative:

The value of `int_(- pi/2)^(pi/2) cos  x  "d"x` is


Choose the correct alternative:

Γ(1) is


`int x^3/(x + 1)` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×