Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\text{We have}, \]
\[I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} x \cos^2 x\ d x\]
\[Let f\left( x \right) = x \cos^2 x\]
\[ \Rightarrow f\left( - x \right) = \left( - x \right) \cos^2 \left( - x \right)\]
\[ = - x \cos^2 x\]
\[ \therefore f\left( - x \right) = - f\left( x \right)\]
\[i . e . , f\left( x \right) \text{is odd function}\]
\[\text{We know that} \int_{- a}^a f\left( x \right) d x = 0 , \text{if }f\left( x \right) \text{is odd function} . \]
\[ \therefore I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} x \cos^2 x\ d x = 0\]
APPEARS IN
संबंधित प्रश्न
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
What is the meaning of an indefinite integral?
