हिंदी

15 ∫ 0 [ X 2 ] D X

Advertisements
Advertisements

प्रश्न

\[\int\limits_0^{15} \left[ x^2 \right] dx\]

योग
Advertisements

उत्तर

We have,

\[I = \int\limits_0^{1 . 5} \left[ x^2 \right] dx\]

\[ = \int\limits_0^1 \left[ x^2 \right] dx + \int\limits_1^\sqrt{2} \left[ x^2 \right] dx + \int\limits_\sqrt{2}^{1 . 5} \left[ x^2 \right] dx\]

\[ = \int\limits_0^1 \left( 0 \right) dx + \int\limits_1^\sqrt{2} \left( 1 \right) dx + \int\limits_\sqrt{2}^{1 . 5} \left( 2 \right) dx ..............\left(\because \left[ x^2 \right] = \begin{cases}0 &where,& 0 < x < 1 \\ 1 &where,& 1 < x < \sqrt{2}\\2 &where,& \sqrt{2} < x < 1.5 \end{cases}\right)\]

\[ = 0 + \left[ x \right]_1^\sqrt{2} + \left[ 2x \right]_\sqrt{2}^{1 . 5} \]

\[ = \left[ x \right]_1^\sqrt{2} + 2 \left[ x \right]_\sqrt{2}^{1 . 5} \]

\[ = \left( \sqrt{2} - 1 \right) + 2\left( 1 . 5 - \sqrt{2} \right)\]

\[ = \sqrt{2} - 1 + 3 - 2\sqrt{2}\]

\[ = 2 - \sqrt{2}\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - Revision Exercise [पृष्ठ १२२]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Revision Exercise | Q 45 | पृष्ठ १२२

संबंधित प्रश्न

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

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

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

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

Evaluate the following definite integrals:

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

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

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

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

\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]


Evaluate each of the following integral:

\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]


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

 


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

\[\int\limits_a^b e^x dx\]

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

\[\int\limits_0^4 \frac{1}{\sqrt{16 - x^2}} dx .\]

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

\[\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{a - \sin \theta}{a + \sin \theta} \right) d\theta\]

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

Evaluate each of the following integral:

\[\int_e^{e^2} \frac{1}{x\log x}dx\]

The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\]  is 


\[\int\limits_0^\pi \frac{1}{a + b \cos x} dx =\]

\[\int\limits_0^1 \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\} dx\] is equal to

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


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


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


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


\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]


\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]


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


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


Evaluate the following:

`int_0^2 "f"(x)  "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`


Evaluate the following:

Γ(4)


Evaluate the following integrals as the limit of the sum:

`int_0^1 x^2  "d"x`


Choose the correct alternative:

If n > 0, then Γ(n) is


Choose the correct alternative:

`int_0^oo x^4"e"^-x  "d"x` is


If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.


Verify the following:

`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`


The value of `int_2^3 x/(x^2 + 1)`dx is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×