हिंदी

9 ∫ 4 √ X ( 30 − X 3 / 2 ) 2 D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\limits_4^9 \frac{\sqrt{x}}{\left( 30 - x^{3/2} \right)^2} dx\]

उत्तर

\[Let\ I = \int_4^9 \frac{\sqrt{x}}{\left( 30 - x^\frac{3}{2} \right)^2} d x . Then, \]
\[Let \left( 30 - x^\frac{3}{2} \right) = t . Then, - \frac{3}{2}\sqrt{x} dx = dt\]
\[When\, x = 4, t = 22\ and\ x\ = 9, t = 3\]
\[ \therefore I = \int_{22}^3 - \frac{2}{3}\frac{1}{t^2} dt\]
\[ \Rightarrow I = \frac{2}{3} \left[ \frac{1}{t} \right]_{22}^3 \]
\[ \Rightarrow I = \frac{2}{3}\left( \frac{1}{3} - \frac{1}{22} \right)\]
\[ \Rightarrow I = \frac{19}{99}\]

shaalaa.com
Definite Integrals
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 20: Definite Integrals - Exercise 20.2 [पृष्ठ ४०]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Exercise 20.2 | Q 47 | पृष्ठ ४०

संबंधित प्रश्न

\[\int\limits_{- 2}^3 \frac{1}{x + 7} dx\]

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

\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]


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

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

\[\int\limits_1^3 \frac{\cos \left( \log x \right)}{x} dx\]

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

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

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

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


\[\int_{- \frac{\pi}{4}}^\frac{\pi}{2} \sin x\left| \sin x \right|dx\]

 


\[\int_{- \frac{\pi}{2}}^\pi \sin^{- 1} \left( \sin x \right)dx\]

\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{- \frac{\pi}{2}}{\sqrt{\cos x \sin^2 x}}dx\]

Evaluate each of the following integral:

\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]


\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\tan x}} dx\]

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

\[\int\limits_{- \pi/4}^{\pi/4} \sin^2 x\ dx\]

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

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

 


\[\int\limits_0^2 \left( x^2 + 4 \right) dx\]

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

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

\[\int\limits_{- 2}^1 \frac{\left| x \right|}{x} dx .\]

\[\int\limits_0^{\pi/2} \log \left( \frac{3 + 5 \cos x}{3 + 5 \sin x} \right) dx .\]

 


Solve each of the following integral:

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

\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals


\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\cot}x} dx\] is

\[\int\limits_0^3 \frac{3x + 1}{x^2 + 9} dx =\]

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

 


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


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


\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]


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


Find : `∫_a^b logx/x` dx


Using second fundamental theorem, evaluate the following:

`int_0^1 x"e"^(x^2)  "d"x`


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


The value of `int_2^3 x/(x^2 + 1)`dx is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×