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.

 

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

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

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

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

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

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

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

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

Evaluate the following integral:

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

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

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

If `f` is an integrable function such that f(2a − x) = f(x), then prove that

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

 


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

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

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

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

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

\[\int\limits_2^3 x^2 dx\]

Evaluate each of the following integral:

\[\int_e^{e^2} \frac{1}{x\log x}dx\]

Evaluate each of the following integral:

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

 


Solve each of the following integral:

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

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

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

 


The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is 

 


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

\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]


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


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


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


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


\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]


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


Using second fundamental theorem, evaluate the following:

`int_1^2 (x "d"x)/(x^2 + 1)`


Using second fundamental theorem, evaluate the following:

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


Evaluate the following using properties of definite integral:

`int_(-1)^1 log ((2 - x)/(2 + x))  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_0^1 (x + 4)  "d"x`


Choose the correct alternative:

`int_(-1)^1 x^3 "e"^(x^4)  "d"x` is


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


Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1


If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.


Evaluate the following:

`int ((x^2 + 2))/(x + 1) "d"x`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×