मराठी

If x = dtt∫0ydt1+9t2 and ddd2ydx2 = ay, then a equal to ______. - Mathematics

Advertisements
Advertisements

प्रश्न

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

पर्याय

  • 3

  • 6

  • 9

  • 1

MCQ
रिकाम्या जागा भरा
Advertisements

उत्तर

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

Explanation:

Since x = `int_0^y "dt"/sqrt(1 + 9"t"^2)`

⇒ `"dx"/"dy" = 1/sqrt(1 + 9y^2)`

which gives `("d"^2y)/("dx"^2) = (18y)/(2sqrt(1 + 9y^2)) * "dy"/"dx"`

= 9y.

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Integrals - Solved Examples [पृष्ठ १६१]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
पाठ 7 Integrals
Solved Examples | Q 25 | पृष्ठ १६१

संबंधित प्रश्‍न

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

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

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

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

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

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

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

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

\[\int_0^\frac{\pi}{4} \frac{\sin^2 x \cos^2 x}{\left( \sin^3 x + \cos^3 x \right)^2}dx\]

\[\int\limits_0^9 f\left( x \right) dx, where f\left( x \right) \begin{cases}\sin x & , & 0 \leq x \leq \pi/2 \\ 1 & , & \pi/2 \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9\end{cases}\]

Evaluate the following integral:

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

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

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

 


\[\int\limits_0^\pi x \sin x \cos^4 x\ dx\]

\[\int\limits_0^\pi \log\left( 1 - \cos x \right) dx\]

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

\[\int\limits_3^5 \left( 2 - x \right) dx\]

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

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

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

\[\int\limits_a^b \frac{f\left( x \right)}{f\left( x \right) + f\left( a + b - x \right)} dx .\]

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

The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .


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


\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{\sin 2x} dx\]  is equal to

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

\[\int\limits_0^\infty \log\left( x + \frac{1}{x} \right) \frac{1}{1 + x^2} dx =\] 

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


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


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


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


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


Using second fundamental theorem, evaluate the following:

`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7)  "d"x`


Evaluate the following using properties of definite integral:

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


Evaluate the following using properties of definite integral:

`int_0^1 x/((1 - x)^(3/4))  "d"x`


Evaluate the following:

`int_0^oo "e"^(-4x) x^4  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_0^1 (x + 4)  "d"x`


If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×