हिंदी

1 ∫ − 1 1 1 + X 2 D X

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[Let\ I = \int_{- 1}^1 \frac{1}{1 + x^2} d x . Then, \]
\[I = \left[ \tan^{- 1} x \right]_{- 1}^1 \]
\[ \Rightarrow I = \tan^{- 1} 1 - \tan^{- 1} \left( - 1 \right)\]
\[ \Rightarrow I = \frac{\pi}{4} - \left( - \frac{\pi}{4} \right)\]
\[ \Rightarrow I = \frac{\pi}{2}\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - Exercise 20.1 [पृष्ठ १६]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Exercise 20.1 | Q 7 | पृष्ठ १६

संबंधित प्रश्न

\[\int\limits_4^9 \frac{1}{\sqrt{x}} dx\]

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

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

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

\[\int\limits_0^{2\pi} e^{x/2} \sin\left( \frac{x}{2} + \frac{\pi}{4} \right) dx\]

\[\int\limits_0^{2\pi} e^x \cos\left( \frac{\pi}{4} + \frac{x}{2} \right) dx\]

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

\[\int\limits_0^{\pi/2} \sqrt{\sin \phi} \cos^5 \phi\ d\phi\]

 


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

\[\int\limits_0^1 \frac{24 x^3}{\left( 1 + x^2 \right)^4} dx\]

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

Evaluate the following integral:

\[\int\limits_{- 3}^3 \left| x + 1 \right| dx\]

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

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

 


\[\int\limits_0^3 \left( x + 4 \right) dx\]

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

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

Solve each of the following integral:

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

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

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

The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .


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


Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .


`int_0^(2a)f(x)dx`


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


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


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


Using second fundamental theorem, evaluate the following:

`int_0^(1/4) sqrt(1 - 4)  "d"x`


Using second fundamental theorem, evaluate the following:

`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`


Using second fundamental theorem, evaluate the following:

`int_1^"e" ("d"x)/(x(1 + logx)^3`


Evaluate the following using properties of definite integral:

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


If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`


Evaluate the following integrals as the limit of the sum:

`int_1^3 (2x + 3)  "d"x`


Choose the correct alternative:

`int_0^oo x^4"e"^-x  "d"x` is


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


Verify the following:

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


`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:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×