Advertisements
Advertisements
प्रश्न
Evaluate: `int_-1^2 (|"x"|)/"x"d"x"`.
Advertisements
उत्तर
`|"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
संबंधित प्रश्न
Evaluate :`int_0^(pi/2)1/(1+cosx)dx`
Evaluate :
`∫_(-pi)^pi (cos ax−sin bx)^2 dx`
find `∫_2^4 x/(x^2 + 1)dx`
Evaluate the integral by using substitution.
`int_0^1 sin^(-1) ((2x)/(1+ x^2)) dx`
The value of the integral `int_(1/3)^4 ((x- x^3)^(1/3))/x^4` dx is ______.
`int 1/(1 + cos x)` dx = _____
A) `tan(x/2) + c`
B) `2 tan (x/2) + c`
C) -`cot (x/2) + c`
D) -2 `cot (x/2)` + c
Evaluate of the following integral:
Evaluate:
Evaluate :
Evaluate:
Evaluate the following integral:
\[\int\limits_0^2 \left| x^2 - 3x + 2 \right| dx\]
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate : \[\int\limits_{- 2}^1 \left| x^3 - x \right|dx\] .
Find : \[\int\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\] .
Evaluate: \[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x}dx\] .
If `I_n = int_0^(pi/4) tan^n theta "d"theta " then " I_8 + I_6` equals ______.
`int_0^1 sin^-1 ((2x)/(1 + x^2))"d"x` = ______.
Evaluate the following:
`int "dt"/sqrt(3"t" - 2"t"^2)`
Find: `int (dx)/sqrt(3 - 2x - x^2)`
Evaluate: `int_0^(π/2) sin 2x tan^-1 (sin x) dx`.
