मराठी

3 ∫ 1 ( 2 X + 3 ) D X

Advertisements
Advertisements

प्रश्न

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

उत्तर

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

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

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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Exercise 20.6 [पृष्ठ ११०]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Exercise 20.6 | Q 6 | पृष्ठ ११०

संबंधित प्रश्‍न

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

\[\int\limits_{- \pi/4}^{\pi/4} \frac{1}{1 + \sin x} dx\]

\[\int\limits_1^4 \frac{x^2 + x}{\sqrt{2x + 1}} dx\]

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

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

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

\[\int_0^1 x\log\left( 1 + 2x \right)dx\]

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

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

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

\[\int\limits_0^{\pi/4} \frac{\tan^3 x}{1 + \cos 2x} dx\]

\[\int\limits_{- 1}^1 5 x^4 \sqrt{x^5 + 1} dx\]

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


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

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

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

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

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

 


\[\int\limits_{- \pi/2}^{\pi/2} \sin^3 x\ dx\]

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

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

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

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

Evaluate each of the following integral:

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

 


\[\int\limits_0^1 e^\left\{ x \right\} dx .\]

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


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

If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals


\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{\sin 2x} dx\]  is equal to

The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is

 


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

 


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


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


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


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


If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`


If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.


Which integral has lower and upper limits?


Using \[F\] as an antiderivative, how is \[\int_{a}^{b}f(x)\,dx\] evaluated?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×