English

3 ∫ 1 ( 2 X + 3 ) D X - Mathematics

Advertisements
Advertisements

Question

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

Solution

\[\int_a^b f\left( x \right) d x = \lim_{h \to 0} h\left[ f\left( a \right) + f\left( a + h \right) + f\left( a + 2h \right) . . . . . . . . . . . . . . . + f\left( a + \left( n - 1 \right)h \right) \right]\]
\[\text{where }h = \frac{b - a}{n}\]

\[\text{Here, }a = 1, b = 3, f\left( x \right) = 2x + 3, h = \frac{3 - 1}{n} = \frac{2}{n}\]

Therefore,

\[I = \int_1^3 \left( 2x + 3 \right) d x\]
\[ = \lim_{h \to 0} h\left[ f\left( 1 \right) + f\left( 1 + h \right) + . . . . . . . . . . . . . . . . . . . . + f\left\{ 1 + \left( n - 1 \right)h \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ \left( 2 + 3 \right) + \left( 2 + 2h + 3 \right) + . . . . . . . . . . . . . . . + \left\{ 2 + 2\left( n - 1 \right)h + 3 \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ 5n + 2h\left\{ 1 + 2 + 3 . . . . . . . . . + \left( n - 1 \right) \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ 5n + 2h\frac{n\left( n - 1 \right)}{2} \right]\]
\[ = \lim_{n \to \infty} \frac{2}{n}\left[ 5n + 2n - 2 \right]\]
\[ = \lim_{n \to \infty} 2\left( 7 - \frac{2}{n} \right)\]
\[ = 14\]

shaalaa.com
Definite Integrals
  Is there an error in this question or solution?
Chapter 20: Definite Integrals - Exercise 20.6 [Page 110]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 20 Definite Integrals
Exercise 20.6 | Q 6 | Page 110

RELATED QUESTIONS

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

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

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

\[\int\limits_{\pi/6}^{\pi/4} cosec\ x\ dx\]

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

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

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

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

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

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

\[\int\limits_4^{12} x \left( x - 4 \right)^{1/3} dx\]

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

\[\int\limits_0^9 f\left( x \right) dx, where f\left( x \right) \begin{cases}\sin x & , & 0 \leq x \leq \pi/2 \\ 1 & , & \pi/2 \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9\end{cases}\]

Evaluate the following integral:

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

\[\int_{- 2}^2 x e^\left| x \right| dx\]

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

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

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

If f(2a − x) = −f(x), prove that

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

Prove that:

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

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

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

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

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

\[\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{a - \sin \theta}{a + \sin \theta} \right) d\theta\]

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

The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is 


\[\int\limits_1^e \log x\ dx =\]

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

\[\int\limits_0^{\pi/2} \sin\ 2x\ \log\ \tan x\ dx\]  is equal to 

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


\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]


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


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


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


Evaluate the following:

`Γ (9/2)`


Choose the correct alternative:

`int_0^oo x^4"e"^-x  "d"x` is


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


`int x^3/(x + 1)` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×