मराठी

3 ∫ 1 ( 2 X 2 + 5 X ) D X

Advertisements
Advertisements

प्रश्न

\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]

बेरीज
Advertisements

उत्तर

\[\text{Here, }a = 1, b = 3, f\left( x \right) = 2 x^2 + 5x, h = \frac{3 - 1}{n} = \frac{2}{n}\]

Therefore,

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

\[ = \lim_{h \to 0} h\left[ f\left( 1 \right) + f\left( 1 + h \right) + . . . . . . . . . . + f\left( 1 + \left( n - 1 \right)h \right) \right]\]

\[ = \lim_{h \to 0} h\left[ 2 + 5 + 2 \left( 1 + h \right)^2 + 5\left( 1 + h \right) + 2 \left( 1 + 2h \right)^2 + 5\left( 1 + 2h \right) + . . . . . . . . . + 2 \left( \left( n - 1 \right)h \right)^2 + 5\left( \left( n - 1 \right)h \right) \right]\]

\[ = \lim_{h \to 0} h\left[ 2n + 2 h^2 \left( 1^2 + 2^2 + . . . . . . . . . . . . . . \left( n - 1 \right)^2 \right) + 4h\left( 1 + 2 + . . . . . . . . . . . . \left( n - 1 \right) \right) + 5n + 5h\left( 1 + 2 + . . . . . . . . . . . . \left( n - 1 \right) \right) \right]\]

\[ = \lim_{h \to 0} h\left[ 7n + 2 h^2 \frac{n\left( n - 1 \right)\left( 2n - 1 \right)}{6} + 9h\frac{n\left( n - 1 \right)}{2} \right]\]

\[ = \lim_{n \to 0 } \left[ 14 + \frac{8}{3}\left( 1 - \frac{1}{n} \right)\left( 2 - \frac{1}{n} \right) + 18\left( 1 - \frac{1}{n} \right) \right]\]

\[ = 14 + \frac{16}{3} + 18\]

\[ = \frac{112}{3}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Revision Exercise [पृष्ठ १२३]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Revision Exercise | Q 66 | पृष्ठ १२३

संबंधित प्रश्‍न

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

\[\int_0^\frac{\pi}{4} \left( a^2 \cos^2 x + b^2 \sin^2 x \right)dx\]

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

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

\[\int\limits_0^{\pi/4} \left( \sqrt{\tan}x + \sqrt{\cot}x \right) dx\]

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

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

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

\[\int_0^\frac{\pi}{4} \frac{\sin^2 x \cos^2 x}{\left( \sin^3 x + \cos^3 x \right)^2}dx\]

\[\int_0^\frac{\pi}{2} \frac{\cos x}{\left( \cos\frac{x}{2} + \sin\frac{x}{2} \right)^n}dx\]

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

 


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

If f is an integrable function, show that

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


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

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

\[\int\limits_0^2 e^x dx\]

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

\[\int\limits_1^4 \left( x^2 - x \right) dx\]

\[\int\limits_{- \pi/2}^{\pi/2} \cos^2 x\ dx .\]

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

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

\[\int\limits_2^3 \frac{1}{x}dx\]

Evaluate each of the following  integral:

\[\int_0^1 x e^{x^2} dx\]

 


Evaluate each of the following integral:

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

 


Evaluate : 

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

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

`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`


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

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

 


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


\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]


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


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


\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]


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


Using second fundamental theorem, evaluate the following:

`int_0^(1/4) sqrt(1 - 4)  "d"x`


Using second fundamental theorem, evaluate the following:

`int_1^2 (x - 1)/x^2  "d"x`


Evaluate the following:

`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`


Evaluate the following using properties of definite integral:

`int_(- pi/2)^(pi/2) sin^2theta  "d"theta`


Evaluate the following using properties of definite integral:

`int_0^1 x/((1 - x)^(3/4))  "d"x`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×