Advertisements
Advertisements
प्रश्न
Evaluate: `int_-1^2 (|"x"|)/"x"d"x"`.
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 (1+logx)/(x(2+logx)(3+logx))dx`
Evaluate :
`∫_0^π(4x sin x)/(1+cos^2 x) dx`
Evaluate the integral by using substitution.
`int_0^(pi/2) sqrt(sin phi) cos^5 phidphi`
Evaluate the integral by using substitution.
`int_0^2 xsqrt(x+2)` (Put x + 2 = `t^2`)
Evaluate the integral by using substitution.
`int_(-1)^1 dx/(x^2 + 2x + 5)`
Evaluate the integral by using substitution.
`int_1^2 (1/x- 1/(2x^2))e^(2x) dx`
Evaluate of the following integral:
Evaluate:
Evaluate:
Evaluate the following integral:
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 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 the following integral:
Evaluate the following integral:
Evaluate :
Find : \[\int\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\] .
Evaluate: \[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x}dx\] .
Evaluate: `int_ e^x ((2+sin2x))/cos^2 x dx`
`int_(pi/5)^((3pi)/10) [(tan x)/(tan x + cot x)]`dx = ?
`int_0^(pi4) sec^4x "d"x` = ______.
Evaluate: `int_0^(π/2) sin 2x tan^-1 (sin x) dx`.
Evaluate: `int x/(x^2 + 1)"d"x`
