English

Evaluate : 3 ∫ 2 3 X D X .

Advertisements
Advertisements

Question

Evaluate : 

\[\int\limits_2^3 3^x dx .\]
Sum
Advertisements

Solution

\[I = \int_2^3 3^x \]
\[ = \left.\frac{3^x}{\log3}\right|_2^3 + C .............\left(Use: \int a^x = \frac{a^x}{\log a} + C\right)\]
\[ = \frac{3^3}{\log3} - \frac{3^2}{\log3} + C\]

\[= \frac{1}{\log3}( 3^3 - 3^2 ) + C\]
\[ = \frac{1}{\log3}(27 - 9) + C\]
\[ = \frac{1}{\log3}(18) + C\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Definite Integrals - Very Short Answers [Page 116]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Very Short Answers | Q 36 | Page 116

RELATED QUESTIONS

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

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

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

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

\[\int\limits_1^2 \frac{3x}{9 x^2 - 1} dx\]

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

 


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

\[\int_0^\frac{1}{2} \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\]

\[\int\limits_0^1 \frac{1 - x^2}{x^4 + x^2 + 1} dx\]

\[\int\limits_0^{\pi/4} \sin^3 2t \cos 2t\ dt\]

\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]


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

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

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

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

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

\[\int\limits_0^\pi x \sin^3 x\ dx\]

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

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

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

\[\int\limits_a^b x\ dx\]

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

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \tan\ xdx\]

 


The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\]  is 


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

 


\[\int\limits_1^2 x\sqrt{3x - 2} dx\]


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


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


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


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


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


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


Using second fundamental theorem, evaluate the following:

`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`


Choose the correct alternative:

`int_(-1)^1 x^3 "e"^(x^4)  "d"x` is


Verify the following:

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


What is the meaning of a definite integral?


What is an antiderivative of \[2x+1\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×