मराठी

2 ∫ 0 X [ X ] D X . - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\limits_0^2 x\left[ x \right] dx .\]
बेरीज

उत्तर

\[\text{We have}, \]
\[I = \int\limits_0^2 x\left[ x \right] dx\]
\[\text{We know that}, \]
\[x\left[ x \right] = \begin{cases}x \times 0&,& 0 < x < 1\\x \times 1&,& 1 < x < 2\end{cases}\]
\[i . e . , \]
\[x\left[ x \right] = \begin{cases}0&,& 0 < x < 1\\x&,& 1 < x < 2\end{cases}\]
\[ \therefore I = \int\limits_0^2 x\left[ x \right] dx\]
\[ = \int\limits_0^1 x\left[ x \right] dx + \int\limits_1^2 x\left[ x \right] dx\]
\[ = \int\limits_0^1 \left( 0 \right) dx + \int\limits_1^2 \left( x \right) dx\]
\[ = 0 + \left[ \frac{x^2}{2} \right]_1^2 \]
\[ = \frac{2^2}{2} - \frac{1^2}{2}\]
\[ = \frac{4}{2} - \frac{1}{2}\]
\[ = \frac{3}{2}\]

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - Very Short Answers [पृष्ठ ११६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Very Short Answers | Q 41 | पृष्ठ ११६

संबंधित प्रश्‍न

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

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

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

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

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

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

Evaluate the following definite integrals:

\[\int_0^\frac{\pi}{2} x^2 \sin\ x\ dx\]

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

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

\[\int\limits_0^{\pi/2} \frac{1}{5 + 4 \sin x} dx\]

\[\int\limits_0^{\pi/4} \frac{\tan^3 x}{1 + \cos 2x} dx\]

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

\[\int\limits_0^{\pi/6} \cos^{- 3} 2 \theta \sin 2\ \theta\ d\ \theta\]

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

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

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

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

\[\int\limits_0^\pi \log\left( 1 - \cos x \right) dx\]

If f is an integrable function, show that

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

 


\[\int\limits_2^3 \left( 2 x^2 + 1 \right) dx\]

\[\int\limits_0^2 \left( x^2 + 4 \right) dx\]

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

If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:

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

 


\[\int\limits_0^{\pi/2} \frac{\cos x}{\left( 2 + \sin x \right)\left( 1 + \sin x \right)} dx\] equals

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

 


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

Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .


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


\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]


\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]


\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]


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


\[\int\limits_0^\pi \cos 2x \log \sin x dx\]


\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]


Evaluate the following:

f(x) = `{{:("c"x",", 0 < x < 1),(0",",  "otherwise"):}` Find 'c" if `int_0^1 "f"(x)  "d"x` = 2


Evaluate the following using properties of definite integral:

`int_(- pi/2)^(pi/2) sin^2theta  "d"theta`


Choose the correct alternative:

Γ(n) is


Evaluate the following:

`int ((x^2 + 2))/(x + 1) "d"x`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×