Advertisements
Advertisements
प्रश्न
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
Advertisements
उत्तर
We have,
\[\left| \sin2\pi x \right| = \begin{cases}\left( \sin2\pi x \right),& 0 \leq x \leq \frac{1}{2}\\ - \left( \sin2\pi x \right),& \frac{1}{2} \leq x \leq 1\end{cases}\]
\[ \therefore \int_0^1 \left| \sin2\pi x \right| d x = \int_0^\frac{1}{2} \sin2\pi x dx + \int_\frac{1}{2}^1 - \sin2\pi x dx\]
\[ = \left[ \frac{- \cos2\pi x}{2\pi} \right]_0^\frac{1}{2} + \left[ \frac{\cos2\pi x}{2\pi} \right]_\frac{1}{2}^1 \]
\[ = \frac{1}{2\pi} + \frac{1}{2\pi} + \frac{1}{2\pi} + \frac{1}{2\pi}\]
\[ = \frac{2}{\pi}\]
APPEARS IN
संबंधित प्रश्न
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
Evaluate the following using properties of definite integral:
`int_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
Choose the correct alternative:
If n > 0, then Γ(n) is
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
What does a definite integral represent?
What is the meaning of an indefinite integral?
