English

√ 2 ∫ 0 [ X 2 ] D X . - Mathematics

Advertisements
Advertisements

Question

\[\int\limits_0^\sqrt{2} \left[ x^2 \right] dx .\]
Sum

Solution

\[\text{We have}, \]
\[I = \int\limits_0^\sqrt{2} \left[ x^2 \right] dx\]
\[ = \int\limits_0^1 \left[ x^2 \right] dx + \int\limits_1^\sqrt{2} \left[ x^2 \right] dx\]
\[ = \int\limits_0^1 \left( 0 \right)dx + \int\limits_1^\sqrt{2} \left( 1 \right)dx .................\left( \because \left[ x^2 \right] = \begin{cases}0&& 0 < x < 1\\1&& 1 < x < \sqrt{2}\end{cases} \right)\]
\[ = 0 + \left[ x \right]_1^\sqrt{2} \]
\[ = \sqrt{2} - 1\]

shaalaa.com
Definite Integrals
  Is there an error in this question or solution?
Chapter 20: Definite Integrals - Very Short Answers [Page 116]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 20 Definite Integrals
Very Short Answers | Q 44 | Page 116

RELATED QUESTIONS

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

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

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

\[\int\limits_1^3 \frac{\log x}{\left( x + 1 \right)^2} dx\]

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

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

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

\[\int_0^\frac{1}{2} \frac{1}{\left( 1 + x^2 \right)\sqrt{1 - x^2}}dx\]

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

\[\int\limits_0^9 f\left( x \right) dx, where f\left( x \right) \begin{cases}\sin x & , & 0 \leq x \leq \pi/2 \\ 1 & , & \pi/2 \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9\end{cases}\]

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

 


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

 


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

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

If f (x) is a continuous function defined on [0, 2a]. Then, prove that

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

 


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

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

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

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

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

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \tan\ xdx\]

 


Evaluate each of the following  integral:

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

 


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

\[\int\limits_0^3 \frac{3x + 1}{x^2 + 9} dx =\]

\[\lim_{n \to \infty} \left\{ \frac{1}{2n + 1} + \frac{1}{2n + 2} + . . . + \frac{1}{2n + n} \right\}\] is equal to

Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .


\[\int\limits_0^4 x\sqrt{4 - x} dx\]


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


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


\[\int\limits_2^3 e^{- x} dx\]


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


Using second fundamental theorem, evaluate the following:

`int_0^1 x"e"^(x^2)  "d"x`


Evaluate the following using properties of definite integral:

`int_(- pi/4)^(pi/4) x^3 cos^3 x  "d"x`


Evaluate the following using properties of definite integral:

`int_0^1 log (1/x - 1)  "d"x`


Evaluate the following:

`int_0^oo "e"^(-4x) x^4  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_0^1 x^2  "d"x`


Choose the correct alternative:

`int_0^oo "e"^(-2x)  "d"x` is


Choose the correct alternative:

Γ(1) is


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


Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×