मराठी

1 ∫ 0 1 X 2 + 1 D X

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\int_0^1 \frac{1}{1 + x^2} d x\]

\[ = \left[ \tan^{- 1} x \right]_0^1 \]

\[ = \frac{\pi}{4} - 0\]

\[ = \frac{\pi}{4}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Very Short Answers [पृष्ठ ११५]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Very Short Answers | Q 8 | पृष्ठ ११५

संबंधित प्रश्‍न

\[\int\limits_2^3 \frac{x}{x^2 + 1} dx\]

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

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

 


\[\int\limits_1^2 \log\ x\ dx\]

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

\[\int\limits_0^2 \frac{1}{\sqrt{3 + 2x - x^2}} dx\]

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

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

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

\[\int\limits_2^4 \frac{x}{x^2 + 1} dx\]

\[\int\limits_0^{\pi/4} \frac{\tan^3 x}{1 + \cos 2x} dx\]

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


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

\[\int_0^{2\pi} \cos^{- 1} \left( \cos x \right)dx\]

Evaluate each of the following integral:

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

 


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

 


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

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

If f (x) is a continuous function defined on [0, 2a]. Then, prove that

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

 


If f(x) is a continuous function defined on [−a, a], then prove that 

\[\int\limits_{- a}^a f\left( x \right) dx = \int\limits_0^a \left\{ f\left( x \right) + f\left( - x \right) \right\} dx\]

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

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

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

\[\int\limits_0^{\pi/4} \tan^2 x\ dx .\]

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

The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is


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


\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]


\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]


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


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


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


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


Evaluate the following using properties of definite integral:

`int_(- pi/2)^(pi/2) sin^2theta  "d"theta`


Evaluate the following:

Γ(4)


Choose the correct alternative:

If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x)  "d"x + int_"c"^"b" f(x)  "d"x` is


Choose the correct alternative:

Γ(1) is


What is the result of a definite integral?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×