Advertisements
Advertisements
Question
Evaluate: `int_-1^2 (|"x"|)/"x"d"x"`.
Advertisements
Solution
`|"x"| = "x" "when" "x" ≥0`
= `-"x" "when" "x" < 0`
Therefore, `|"x"|/"x"` = 1 when x ≥ 0
= -1 when x < 0
Thus, `int_-1^2 |"x"|/"x"d"x" = int_-1^0 (-1)d"x" + int_0^2 (1)d"x"`
= `-1 xx ["x"]_1^0 + ["x"]_0^2`
= `(-1) [0 + 1] + [2 - 0] = -1 + 2 = 1`.
APPEARS IN
RELATED QUESTIONS
Evaluate: `int1/(xlogxlog(logx))dx`
Evaluate : `int_0^4(|x|+|x-2|+|x-4|)dx`
If `int_0^a1/(4+x^2)dx=pi/8` , find the value of a.
Evaluate the integral by using substitution.
`int_0^1 x/(x^2 +1)`dx
Evaluate the integral by using substitution.
`int_0^(pi/2) (sin x)/(1+ cos^2 x) dx`
The value of the integral `int_(1/3)^4 ((x- x^3)^(1/3))/x^4` dx is ______.
Evaluate of the following integral:
Evaluate of the following integral:
Evaluate of the following integral:
Evaluate of the following integral:
Evaluate of the following integral:
Evaluate:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate each of the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate
\[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\]
Evaluate :
Evaluate: `int_ e^x ((2+sin2x))/cos^2 x dx`
`int_0^1 sin^-1 ((2x)/(1 + x^2))"d"x` = ______.
Evaluate the following:
`int ("e"^(6logx) - "e"^(5logx))/("e"^(4logx) - "e"^(3logx)) "d"x`
The value of `int_0^1 (x^4(1 - x)^4)/(1 + x^2) dx` is
Evaluate: `int x/(x^2 + 1)"d"x`
