मराठी

9 ∫ 4 1 √ X D X

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[Let I = \int_4^9 \frac{1}{\sqrt{x}} d x . Then, \]
\[I = 2 \int_4^9 \frac{1}{2\sqrt{x}} d x\]
\[ \Rightarrow I = 2 \left[ \sqrt{x} \right]_4^9 \]
\[ \Rightarrow I = 2\left( 3 - 2 \right)\]
\[ \Rightarrow I = 2\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Exercise 20.1 [पृष्ठ १६]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Exercise 20.1 | Q 1 | पृष्ठ १६

संबंधित प्रश्‍न

\[\int\limits_0^{\pi/6} \cos x \cos 2x\ dx\]

Evaluate the following definite integrals:

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

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

\[\int\limits_0^{\pi/2} \sqrt{\sin \phi} \cos^5 \phi\ d\phi\]

 


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

\[\int_0^\frac{\pi}{2} \frac{\cos^2 x}{1 + 3 \sin^2 x}dx\]

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

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

\[\int_\frac{1}{3}^1 \frac{\left( x - x^3 \right)^\frac{1}{3}}{x^4}dx\]

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

 


Evaluate the following integral:

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

 


If f(2a − x) = −f(x), prove that

\[\int\limits_0^{2a} f\left( x \right) dx = 0 .\]

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

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

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

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

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

\[\int\limits_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} dx, n \in N .\]

\[\int\limits_0^\pi \cos^5 x\ dx .\]

\[\int\limits_0^2 \sqrt{4 - x^2} dx\]

If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.


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


\[\int\limits_1^\sqrt{3} \frac{1}{1 + x^2} dx\]  is equal to ______.

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

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

The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is 

 


Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .

 

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


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


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


\[\int\limits_0^{2\pi} \cos^7 x dx\]


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


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


\[\int\limits_0^4 x dx\]


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


Verify the following:

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


Evaluate \[\int_{1}^{3}(2x+1)\,dx\].


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×