English

2 ∫ 0 ( 2 X 2 + 3 ) D X

Advertisements
Advertisements

Question

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

Sum
Advertisements

Solution

\[\int_0^2 \left( 2 x^2 + 3 \right) d x\]
\[ = \left[ \frac{2 x^3}{3} + 3x \right]_0^2 \]
\[ = \frac{16}{3} + 6 = \frac{34}{3}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Definite Integrals - Revision Exercise [Page 123]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Revision Exercise | Q 62 | Page 123

RELATED QUESTIONS

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

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

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

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

\[\int\limits_0^2 \frac{1}{\sqrt{3 + 2x - x^2}} dx\]

\[\int\limits_0^{2\pi} e^{x/2} \sin\left( \frac{x}{2} + \frac{\pi}{4} \right) dx\]

\[\int_0^\frac{\pi}{4} \left( \tan x + \cot x \right)^{- 2} dx\]

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

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

\[\int_0^\frac{1}{2} \frac{1}{\left( 1 + x^2 \right)\sqrt{1 - x^2}}dx\]

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

 


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

\[\int\limits_0^\pi x \log \sin x\ dx\]

\[\int\limits_0^\pi x \cos^2 x\ dx\]

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

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

 


Prove that:

\[\int_0^\pi xf\left( \sin x \right)dx = \frac{\pi}{2} \int_0^\pi f\left( \sin x \right)dx\]

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

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

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

Evaluate each of the following integral:

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

\[\int\limits_0^2 \left[ x \right] dx .\]

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

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

 


\[\int\limits_0^1 \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\} dx\] is equal to

The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is 


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


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


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


\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]


\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]


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


\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]


\[\int\limits_2^3 e^{- x} dx\]


Using second fundamental theorem, evaluate the following:

`int_1^2 (x - 1)/x^2  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_0^1 (x + 4)  "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


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


Find `int x^2/(x^4 + 3x^2 + 2) "d"x`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×