हिंदी

Π ∫ − π X 10 Sin 7 X D X

Advertisements
Advertisements

प्रश्न

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

योग
Advertisements

उत्तर

\[\int_{- \pi}^\pi x^{10} \sin^7 x d x\]

\[Let f\left( x \right) = x^{10} \sin^7 x\]

\[\text{Consider }f\left( - x \right) = \left( - x \right)^{10} \sin^7 \left( - x \right) = - x^{10} \sin^7 x = - f\left( x \right)\]

Hence f(x) is an odd function

Therefore

\[ \int_{- \pi}^\pi x^{10} \sin^7 x d x = 0\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - Revision Exercise [पृष्ठ १२२]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Revision Exercise | Q 55 | पृष्ठ १२२

संबंधित प्रश्न

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

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

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

\[\int\limits_1^e \frac{\log x}{x} dx\]

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

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

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

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

\[\int\limits_0^{\pi/2} \cos^5 x\ dx\]

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

Evaluate the following integral:

\[\int_{- a}^a \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right)d\theta\]

If f is an integrable function, show that

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

 


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

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

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

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

\[\int\limits_0^{\pi/2} \sqrt{1 - \cos 2x}\ dx .\]

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

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

The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is

 


The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is 

 


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

The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is


Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .


\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]


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


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


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


Using second fundamental theorem, evaluate the following:

`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7)  "d"x`


Evaluate the following:

f(x) = `{{:("c"x",", 0 < x < 1),(0",",  "otherwise"):}` Find 'c" if `int_0^1 "f"(x)  "d"x` = 2


Evaluate the following using properties of definite integral:

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


Evaluate the following using properties of definite integral:

`int_0^1 log (1/x - 1)  "d"x`


Evaluate the following:

`int_0^oo "e"^(-4x) x^4  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_0^1 (x + 4)  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_1^3 x  "d"x`


Choose the correct alternative:

If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x)  "d"x + int_"c"^"b" f(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×