हिंदी

If a ∫ 0 1 1 + 4 X 2 D X = π 8 , Then a Equals,π 2,1 2,π 4,1

Advertisements
Advertisements

प्रश्न

If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals

 

विकल्प

  • \[\frac{\pi}{2}\]
  • \[\frac{1}{2}\]
  • \[\frac{\pi}{4}\]
  • 1

MCQ
Advertisements

उत्तर

\[\frac{1}{2}\]

\[\int_0^\alpha \frac{1}{1 + 4 x^2} d x = \frac{\pi}{8}\]

\[ \Rightarrow \int_0^\alpha \frac{1}{1 + \left( 2x \right)^2} d x = \frac{\pi}{8}\]

\[ \Rightarrow \frac{1}{2} \left[ \tan^{- 1} 2x \right]_0^\alpha = \frac{\pi}{8}\]

\[ \Rightarrow \frac{1}{2} \tan^{- 1} 2\alpha = \frac{\pi}{8}\]

\[ \Rightarrow 2\alpha = \tan\frac{\pi}{4}\]

\[ \Rightarrow 2\alpha = 1\]

\[ \therefore \alpha = \frac{1}{2}\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - MCQ [पृष्ठ ११८]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
MCQ | Q 21 | पृष्ठ ११८

संबंधित प्रश्न

\[\int\limits_{\pi/3}^{\pi/4} \left( \tan x + \cot x \right)^2 dx\]

\[\int\limits_1^e \frac{\log x}{x} dx\]

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

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

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

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

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

\[\int\limits_{- 1}^1 5 x^4 \sqrt{x^5 + 1} dx\]

Evaluate the following integral:

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

\[\int_0^\pi \cos x\left| \cos x \right|dx\]

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

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

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

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

Evaluate the following integral:

\[\int_{- 1}^1 \left| xcos\pi x \right|dx\]

 


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

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

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

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

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

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

Evaluate each of the following  integral:

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

 


Write the coefficient abc of which the value of the integral

\[\int\limits_{- 3}^3 \left( a x^2 + bx + c \right) dx\] is independent.

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

\[\int\limits_1^\sqrt{3} \frac{1}{1 + x^2} dx\]  is equal to ______.

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

\[\int\limits_0^1 \frac{x}{\left( 1 - x \right)^\frac{5}{4}} dx =\]

\[\lim_{n \to \infty} \left\{ \frac{1}{2n + 1} + \frac{1}{2n + 2} + . . . + \frac{1}{2n + n} \right\}\] is equal to

The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is

 


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


\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]


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


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


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


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


Evaluate the following using properties of definite integral:

`int_(- pi/4)^(pi/4) x^3 cos^3 x  "d"x`


Choose the correct alternative:

`Γ(3/2)`


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×