Advertisements
Advertisements
प्रश्न
पर्याय
1
2
− 1
− 2
उत्तर
2
\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \sin\left| x \right| d x\]
\[ = - \int_{- \frac{\pi}{2}}^0 \sin x\ dx + \int_0^\frac{\pi}{2} \sin x\ dx\]
\[ = - \left[ - \cos x \right]_{- \frac{\pi}{2}}^0 + \left[ - \cos x \right]_0^\frac{\pi}{2} \]
\[ = 1 - 0 - 0 + 1\]
\[ = 2\]
APPEARS IN
संबंधित प्रश्न
Prove that:
Evaluate each of the following integral:
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following:
Γ(4)
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Find `int sqrt(10 - 4x + 4x^2) "d"x`
