हिंदी

Π / 4 ∫ 0 E X Sin X D X

Advertisements
Advertisements

प्रश्न

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

योग
Advertisements

उत्तर

\[Let, I = \int_0^\frac{\pi}{4} e^x \sin x d x ..............(1)\]
\[ = \left[ - e^x \cos x \right]_0^\frac{\pi}{4} + \int_0^\frac{\pi}{4} e^x \cos x dx\]
\[ = \left[ - e^x \cos x \right]_0^\frac{\pi}{4} + \left[ e^x \sin x \right]_0^\frac{\pi}{4} - \int_0^\frac{\pi}{4} e^x \sin x dx\]
\[ \Rightarrow I = \left[ - e^x \cos x \right]_0^\frac{\pi}{4} + \left[ e^x \sin x \right]_0^\frac{\pi}{4} - I ..............\left[\text{Using (1)} \right] \]
\[ \Rightarrow 2I = \left[ - e^x \cos x \right]_0^\frac{\pi}{4} + \left[ e^x \sin x \right]_0^\frac{\pi}{4} \]
\[ = - \frac{1}{\sqrt{2}} e^\frac{\pi}{4} + 1 + \frac{1}{\sqrt{2}} e^\frac{\pi}{4} - 0\]
\[ = 1\]
\[\text{Hence }I = \frac{1}{2}\]

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 27 | पृष्ठ १२२

संबंधित प्रश्न

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

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

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

\[\int\limits_{\pi/6}^{\pi/4} cosec\ x\ dx\]

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

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

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

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

\[\int\limits_0^\pi 5 \left( 5 - 4 \cos \theta \right)^{1/4} \sin \theta\ d \theta\]

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

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

Evaluate the following integral:

\[\int\limits_{- 2}^2 \left| 2x + 3 \right| dx\]

Evaluate each of the following integral:

\[\int_0^{2\pi} \log\left( \sec x + \tan x \right)dx\]

 


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

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

\[\int_0^1 | x\sin \pi x | dx\]

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

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

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

Evaluate each of the following integral:

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

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

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

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

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


Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]


\[\int\limits_{- 1}^1 \left| 1 - x \right| dx\]  is equal to

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

Evaluate the following integrals :-

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


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


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


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


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


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


Find : `∫_a^b logx/x` dx


Evaluate the following using properties of definite integral:

`int_(-1)^1 log ((2 - x)/(2 + x))  "d"x`


Evaluate the following using properties of definite integral:

`int_0^1 log (1/x - 1)  "d"x`


Evaluate the following:

Γ(4)


Evaluate the following integrals as the limit of the sum:

`int_0^1 (x + 4)  "d"x`


`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.


What is the meaning of a definite integral?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×