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:
\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]
Evaluate each of the following integral:
Solve each of the following integral:
Write the coefficient a, b, c of which the value of the integral
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
