Advertisements
Advertisements
प्रश्न
Evaluate each of the following integral:
Advertisements
उत्तर
\[\int_0^\frac{\pi}{2} e^x \left( \sin x - \cos x \right)dx\]
\[ = - \int_0^\frac{\pi}{2} e^x \left[ \cos x + \left( - \sin x \right) \right]dx\]
\[ = \left.- {e^x \cos x}\right|_0^\frac{\pi}{2} .............\left\{ \int e^x \left[ f\left( x \right) + f'\left( x \right) \right]dx = e^x f\left( x \right) + C \right\}\]
\[ = - \left( e^\frac{\pi}{2} \cos\frac{\pi}{2} - e^0 \cos0 \right)\]
\[ = - \left( e^\frac{\pi}{2} \times 0 - 1 \times 1 \right)\]
\[ = - \left( 0 - 1 \right)\]
\[ = 1\]
संबंधित प्रश्न
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
Evaluate :
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Evaluate the following:
`int_0^2 "f"(x) "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`
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_(- pi/2)^(pi/2) sin^2theta "d"theta`
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
