मराठी

4 ∫ 1 F ( X ) D X , W H E R E F ( X ) = ( 4 X + 3 , I F 1 ≤ X ≤ 2 3 X + 5 , I F 2 ≤ X ≤ 4 ) - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

We have,

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

\[I = \int_0^9 f\left( x \right) d x\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} f\left( x \right) d x + \int_\frac{\pi}{2}^3 f\left( x \right) d x + \int_3^9 f\left( x \right) d x ....................\left[\text{Additive property} \right]\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \sin x d x + \int_\frac{\pi}{2}^3 1 d x + \int_3^9 e^{x - 3} d x\]
\[ \Rightarrow I = \left[ - \cos x \right]_0^\frac{\pi}{2} + \left[ x \right]_\frac{\pi}{2}^3 + \left[ e^{x - 3} \right]_3^9 \]
\[ \Rightarrow I = 0 + 1 + 3 - \frac{\pi}{2} + e^6 - e^0 \]
\[ \Rightarrow I = 3 - \frac{\pi}{2} + e^6\]

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - Exercise 20.3 [पृष्ठ ५५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Exercise 20.3 | Q 1.2 | पृष्ठ ५५

संबंधित प्रश्‍न

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

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

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

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

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

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

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

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

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

\[\int_0^\frac{\pi}{2} \frac{\cos^2 x}{1 + 3 \sin^2 x}dx\]

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

\[\int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx\]

 


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

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

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

 


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

 


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

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

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

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

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

Evaluate each of the following  integral:

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

 


Evaluate each of the following integral:

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

 


If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.


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

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

\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\]  equals


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


\[\int\limits_{- 1}^1 \left| 1 - x \right| dx\]  is equal to

If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\]  then the value of I10 + 90I8 is

 


\[\int\limits_0^1 \frac{x}{\left( 1 - x \right)^\frac{5}{4}} dx =\]

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

If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to


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


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


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


Choose the correct alternative:

`Γ(3/2)`


Verify the following:

`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`


`int (x + 3)/(x + 4)^2 "e"^x  "d"x` = ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×