हिंदी

If 1 ∫ 0 ( 3 X 2 + 2 X + K ) D X = 0 , Find the Value of K. - Mathematics

Advertisements
Advertisements

प्रश्न

If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.

 

Advertisements

उत्तर

\[\text{We have}, \]
\[ \int_0^1 \left( 3 x^2 + 2x + k \right) d x = 0\]
\[ \Rightarrow \left[ x^3 + x^2 + kx \right]_0^1 = 0\]
\[ \Rightarrow 1 + 1 + k - 0 = 0\]
\[ \Rightarrow k = - 2\]

shaalaa.com
Definite Integrals
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 20: Definite Integrals - Very Short Answers [पृष्ठ ११६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Very Short Answers | Q 31 | पृष्ठ ११६

संबंधित प्रश्न

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

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

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

\[\int\limits_{- 1}^1 \frac{1}{x^2 + 2x + 5} dx\]

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

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

\[\int\limits_0^{\pi/4} \left( \sqrt{\tan}x + \sqrt{\cot}x \right) dx\]

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

\[\int_0^\frac{\pi}{2} \frac{\tan x}{1 + m^2 \tan^2 x}dx\]

\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]


Evaluate the following integral:

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

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

 


\[\int\limits_0^\pi \frac{x \tan x}{\sec x \ cosec x} dx\]

\[\int\limits_0^\pi x \sin^3 x\ dx\]

Evaluate the following integral:

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

 


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

\[\int\limits_1^2 \left( x^2 - 1 \right) dx\]

\[\int\limits_1^4 \left( x^2 - x \right) dx\]

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

\[\int\limits_1^4 \left( x^2 - x \right) dx\]

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

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

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

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

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

 


Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .


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


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


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


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


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


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


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


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


Evaluate the following using properties of definite integral:

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


Evaluate the following:

`Γ (9/2)`


Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`


Find `int sqrt(10 - 4x + 4x^2)  "d"x`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×