English

9 ∫ 4 √ X ( 30 − X 3 / 2 ) 2 D X

Advertisements
Advertisements

Question

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

Solution

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

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Definite Integrals - Exercise 20.2 [Page 40]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Exercise 20.2 | Q 47 | Page 40

RELATED QUESTIONS

\[\int\limits_4^9 \frac{1}{\sqrt{x}} dx\]

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

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

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

 


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

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

\[\int\limits_0^{\pi/2} \frac{\sin \theta}{\sqrt{1 + \cos \theta}} d\theta\]

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

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

Evaluate the following integral:

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

\[\int_0^{2\pi} \cos^{- 1} \left( \cos x \right)dx\]

\[\int\limits_0^5 \frac{\sqrt[4]{x + 4}}{\sqrt[4]{x + 4} + \sqrt[4]{9 - x}} dx\]

\[\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x} + \sqrt[3]{7} - x} dx\]

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

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

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

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

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

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

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

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

\[\int\limits_0^{\pi/4} \tan^2 x\ dx .\]

\[\int\limits_1^2 \log_e \left[ x \right] dx .\]

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

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


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

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

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


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


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


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


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


\[\int\limits_0^{15} \left[ x^2 \right] dx\]


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


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


Using second fundamental theorem, evaluate the following:

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


Choose the correct alternative:

Using the factorial representation of the gamma function, which of the following is the solution for the gamma function Γ(n) when n = 8 is


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


Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`


Verify the following:

`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×