Advertisements
Advertisements
प्रश्न
विकल्प
`15/16`
`3/16`
`-3/16`
`-16/3`
Advertisements
उत्तर
`-16/3`
`I=int_0^1x/(1-x)^(5/4)dx`
Put, 1 - x = t ⇒ x = 1 - t
⇒ dx = -dt
| x | 0 | 1 |
| t | 1 | 0 |
`I=int_1^0((1-t)(-dt))/t^(5/4)`
`I=int_0^1(1-t)/t^(5/4)dt`
`I=int_0^1(t^(-5/4)-t^(-1/4))dt`
`I=[t^(-1/4)/(-1/4)-t^(3/4)/(3/4)]_0^1`
`I=-4-4/3`
`I=-16/3`
APPEARS IN
संबंधित प्रश्न
Evaluate the following definite integrals:
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^4 x dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
\[\int\limits_2^3 e^{- x} dx\]
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "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 ______.
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
