English

1 ∫ 0 { X } D X , Where {X} Denotes the Fractional Part of X.

Advertisements
Advertisements

Question

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

 
Advertisements

Solution

\[\text{We have}, \]
\[I = \int\limits_0^1 \left\{ x \right\} dx\]
\[\text{We know} \left\{ x \right\} = x, 0 < x < 1\]
\[ \therefore I = \int\limits_0^1 x\ dx\]
\[ = \left[ \frac{x^2}{2} \right]_0^1 \]
\[ = \frac{1}{2} - \frac{0}{2}\]
\[ = \frac{1}{2}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Definite Integrals - Very Short Answers [Page 116]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Very Short Answers | Q 39 | Page 116

RELATED QUESTIONS

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

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

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

\[\int\limits_{- 1}^1 \frac{1}{x^2 + 2x + 5} dx\]

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

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

\[\int_0^\frac{\pi}{4} \left( \tan x + \cot x \right)^{- 2} dx\]

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

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

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

\[\int\limits_{- 1}^1 5 x^4 \sqrt{x^5 + 1} dx\]

\[\int\limits_0^{\pi/2} \cos^5 x\ dx\]

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

\[\int_0^\frac{\pi}{4} \frac{\sin^2 x \cos^2 x}{\left( \sin^3 x + \cos^3 x \right)^2}dx\]

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

\[\int_0^\pi \cos x\left| \cos x \right|dx\]

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

\[\int_{- \frac{\pi}{2}}^\pi \sin^{- 1} \left( \sin x \right)dx\]

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

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

\[\int\limits_0^\pi x \sin x \cos^4 x\ dx\]

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

\[\int\limits_{- 1}^1 \left( x + 3 \right) dx\]

\[\int\limits_0^2 e^x dx\]

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

\[\int\limits_0^{\pi/4} \tan^2 x\ dx .\]

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

Evaluate each of the following integral:

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

 


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

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

 


\[\int\limits_0^\infty \log\left( x + \frac{1}{x} \right) \frac{1}{1 + x^2} dx =\] 

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


\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]


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


\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]


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_(-1)^1 log ((2 - x)/(2 + x))  "d"x`


Evaluate the following:

`int_0^oo "e"^(-4x) x^4  "d"x`


Evaluate \[\int_{0}^{1}x\,dx\].


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×