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

\[\int\limits_0^{1/2} \frac{1}{\sqrt{1 - x^2}} dx\]

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

\[\int\limits_e^{e^2} \left\{ \frac{1}{\log x} - \frac{1}{\left( \log x \right)^2} \right\} dx\]

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

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

\[\int\limits_0^{\pi/2} \frac{1}{5 \cos x + 3 \sin x} dx\]

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

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

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

\[\int\limits_0^{\pi/2} \cos^5 x\ dx\]

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

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

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

If f(2a − x) = −f(x), prove that

\[\int\limits_0^{2a} f\left( x \right) dx = 0 .\]

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

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

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

Evaluate each of the following integral:

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

 


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


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


`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`


\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\cot}x} dx\] is

\[\int\limits_1^e \log x\ dx =\]

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


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

\[\int\limits_0^{\pi/2} \sin\ 2x\ \log\ \tan x\ dx\]  is equal to 

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


\[\int\limits_0^{2\pi} \cos^7 x dx\]


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


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


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


Evaluate the following using properties of definite integral:

`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x)  "d"x`


Choose the correct alternative:

Γ(1) is


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


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


If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.


Which integral has lower and upper limits?


If velocity changes with time, what does a definite integral give over a time interval?


Using \[F\] as an antiderivative, how is \[\int_{a}^{b}f(x)\,dx\] evaluated?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×