मराठी

Π / 2 ∫ − π / 2 X Cos 2 X D X . - Mathematics

Advertisements
Advertisements

प्रश्न

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

 

बेरीज

उत्तर

\[\text{We have}, \]

\[I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} x \cos^2 x\ d x\]

\[Let f\left( x \right) = x \cos^2 x\]

\[ \Rightarrow f\left( - x \right) = \left( - x \right) \cos^2 \left( - x \right)\]

\[ = - x \cos^2 x\]

\[ \therefore f\left( - x \right) = - f\left( x \right)\]

\[i . e . , f\left( x \right) \text{is odd function}\]

\[\text{We know that} \int_{- a}^a f\left( x \right) d x = 0 , \text{if }f\left( x \right) \text{is odd function} . \]

\[ \therefore I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} x \cos^2 x\ d x = 0\]

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - Very Short Answers [पृष्ठ ११५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Very Short Answers | Q 6 | पृष्ठ ११५

संबंधित प्रश्‍न

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

\[\int\limits_2^3 \frac{x}{x^2 + 1} dx\]

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

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

\[\int\limits_0^{\pi/2} \sin^3 x\ dx\]

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

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

\[\int\limits_2^4 \frac{x}{x^2 + 1} dx\]

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

\[\int\limits_4^9 \frac{\sqrt{x}}{\left( 30 - x^{3/2} \right)^2} dx\]

\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]


Evaluate the following integral:

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

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

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

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

 


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

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

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

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

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

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

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

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


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

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


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


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


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


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


Evaluate the following:

`int_(-1)^1 "f"(x)  "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x  < 0):}`


Evaluate the following using properties of definite integral:

`int_(- pi/4)^(pi/4) x^3 cos^3 x  "d"x`


Evaluate the following using properties of definite integral:

`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x)  "d"x`


Evaluate the following:

`Γ (9/2)`


Choose the correct alternative:

`int_0^oo "e"^(-2x)  "d"x` is


Choose the correct alternative:

`int_(-1)^1 x^3 "e"^(x^4)  "d"x` is


Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1


Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`


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


Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×