हिंदी

1 ∫ 0 2 X − [ X ] D X

Advertisements
Advertisements

प्रश्न

\[\int\limits_0^1 2^{x - \left[ x \right]} dx\]
योग
Advertisements

उत्तर

\[\text{We have}, \]
\[I = \int\limits_0^1 2^{x - \left[ x \right]} dx\]
\[ = \int\limits_0^1 2^{x - 0} dx ...............\left( \because \left[ x \right] = 0\text{ where, }0 < x < 1 \right)\]
\[ = \int\limits_0^1 2^x dx\]
\[ = \left[ \frac{2^x}{\log_e 2} \right]_0^1 \]
\[ = \frac{2^1}{\log_e 2} - \frac{2^0}{\log_e 2}\]
\[ = \frac{2}{\log_e 2} - \frac{1}{\log_e 2}\]
\[ = \frac{1}{\log_e 2}\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - Very Short Answers [पृष्ठ ११६]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Very Short Answers | Q 42 | पृष्ठ ११६

संबंधित प्रश्न

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

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

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

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

\[\int_0^\frac{\pi}{2} \frac{\cos^2 x}{1 + 3 \sin^2 x}dx\]

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

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

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

\[\int_0^2 2x\left[ x \right]dx\]

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

 


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

 


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

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

\[\int\limits_0^\pi \cos^5 x\ dx .\]

Evaluate each of the following  integral:

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

 


Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \sin2xdx\]

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

 

\[\int\limits_0^1 e^\left\{ x \right\} dx .\]

\[\int\limits_1^2 \log_e \left[ x \right] dx .\]

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

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


If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals


\[\int\limits_0^{2a} f\left( x \right) dx\]  is equal to


\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]


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


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


Evaluate the following integrals :-

\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]


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


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


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


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


Evaluate the following:

`int_(-1)^1 "f"(x)  "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x  < 0):}`


Choose the correct alternative:

`int_0^oo "e"^(-2x)  "d"x` is


Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x


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


Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:


What is the meaning of an indefinite integral?


What is an antiderivative of \[x\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×