Advertisements
Advertisements
Question
Options
`15/16`
`3/16`
`-3/16`
`-16/3`
Advertisements
Solution
`-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
RELATED QUESTIONS
Evaluate each of the following integral:
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
Find : `∫_a^b logx/x` dx
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(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):}`
Choose the correct alternative:
Γ(1) is
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
Which integral has lower and upper limits?
What is an antiderivative of \[2x+1\]?
