English

If F(X) is a Continuous Function Defined on [−A, A], Then Prove that a ∫ − a F ( X ) D X = a ∫ 0 { F ( X ) + F ( − X ) } D X - Mathematics

Advertisements
Advertisements

Question

If f(x) is a continuous function defined on [−aa], then prove that 

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

Solution

\[Let\ I = \int_{- a}^a f\left( x \right) d x\]
\[\text{By Additive property}\]
\[I = \int_{- a}^0 f\left( x \right) d x + \int_0^a f\left( x \right) d x\]
\[Let x = - t, then\ dx = - dt, \]
\[When\ x = - a, t = a, x = 0, t = 0\]
\[Hence\ \int_{- a}^0 f\left( x \right) d x = - \int_a^0 f\left( - t \right) d t\]
\[ = \int_0^a f\left( - t \right) d t = \int_0^a f\left( - x \right) dx .......................\left( \text{Changing the variable} \right)\]
Therefore,
\[I = \int_0^a f\left( - x \right) d x + \int_0^a f\left( x \right) d x\]
\[ = \int_0^a \left\{ f\left( x \right) + f\left( - x \right) \right\} dx\]
\[\text{Hence, proved} .\]

shaalaa.com
Definite Integrals
  Is there an error in this question or solution?
Chapter 20: Definite Integrals - Exercise 20.5 [Page 96]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 20 Definite Integrals
Exercise 20.5 | Q 48 | Page 96

RELATED QUESTIONS

\[\int\limits_{- 2}^3 \frac{1}{x + 7} dx\]

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

\[\int\limits_0^{\pi/6} \cos x \cos 2x\ dx\]

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

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

\[\int\limits_0^{\pi/2} \frac{x + \sin x}{1 + \cos x} dx\]

\[\int\limits_0^1 x \tan^{- 1} x\ dx\]

\[\int\limits_0^{\pi/2} \sin 2x \tan^{- 1} \left( \sin x \right) dx\]

Evaluate the following integral:

\[\int\limits_{- 2}^2 \left| 2x + 3 \right| dx\]

\[\int_{- \frac{\pi}{4}}^\frac{\pi}{2} \sin x\left| \sin x \right|dx\]

 


\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx\]

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

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

\[\int\limits_{- \pi/2}^{\pi/2} \sin^3 x\ dx\]

If f(2a − x) = −f(x), prove that

\[\int\limits_0^{2a} f\left( x \right) dx = 0 .\]

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

\[\int\limits_0^1 \left( 3 x^2 + 5x \right) dx\]

\[\int\limits_0^{\pi/2} \log \tan x\ dx .\]

\[\int\limits_{- \pi/2}^{\pi/2} \sin\left| x \right| dx\]  is equal to

If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals

 


\[\int\limits_{- 1}^1 \left| 1 - x \right| dx\]  is equal to

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

The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is

 


Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .


\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]


\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]


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


\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]


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


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


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


Find : `∫_a^b logx/x` dx


Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`


Using second fundamental theorem, evaluate the following:

`int_0^1 "e"^(2x)  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_0^1 (x + 4)  "d"x`


Choose the correct alternative:

`int_(-1)^1 x^3 "e"^(x^4)  "d"x` is


Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`


`int "e"^x ((1 - x)/(1 + x^2))^2  "d"x` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×