Advertisements
Advertisements
प्रश्न
Evaluate: `int_0^1 |x| "d"x`
उत्तर
`int_0^1 |x| "d"x - int_0^1 x "d"x` ......[|x| = x, x ≥ 0]
= `[x^2/2]_0^(-1)`
= `1/2(1^2 - 0)`
= `1/2`
APPEARS IN
संबंधित प्रश्न
Evaluate: `int_0^1 (x^2 - 2)/(x^2 + 1).dx`
Evaluate: `int_0^oo xe^-x.dx`
Evaluate: `int_0^π sin^3x (1 + 2cosx)(1 + cosx)^2.dx`
Evaluate the following:
`int_0^a (1)/(x + sqrt(a^2 - x^2)).dx`
`int_0^1 (x^2 - 2)/(x^2 + 1) "d"x` =
Let I1 = `int_"e"^("e"^2) 1/logx "d"x` and I2 = `int_1^2 ("e"^x)/x "d"x` then
`int_0^4 1/sqrt(4x - x^2) "d"x` =
`int_0^(pi/2) log(tanx) "d"x` =
Evaluate: `int_(pi/6)^(pi/3) cosx "d"x`
Evaluate: `int_(- pi/4)^(pi/4) x^3 sin^4x "d"x`
Evaluate: `int_0^(pi/4) sec^2 x "d"x`
Evaluate: `int_1^2 x/(1 + x^2) "d"x`
Evaluate: `int_0^1 "e"^x/sqrt("e"^x - 1) "d"x`
Evaluate: `int_0^1(x + 1)^2 "d"x`
Evaluate: `int_0^(pi/2) (sin^4x)/(sin^4x + cos^4x) "d"x`
Evaluate: `int_3^8 (11 - x)^2/(x^2 + (11 - x)^2) "d"x`
Evaluate: `int_(-1)^1 |5x - 3| "d"x`
Evaluate: `int_(-4)^2 1/(x^2 + 4x + 13) "d"x`
Evaluate: `int_0^1 x* tan^-1x "d"x`
Evaluate: `int_(-1)^1 1/("a"^2"e"^x + "b"^2"e"^(-x)) "d"x`
Evaluate: `int_0^3 x^2 (3 - x)^(5/2) "d"x`
Evaluate: `int_0^(pi/4) (sec^2x)/(3tan^2x + 4tan x + 1) "d"x`
Evaluate: `int_0^1 (log(x + 1))/(x^2 + 1) "d"x`
Evaluate: `int_0^pi x*sinx*cos^2x* "d"x`
Evaluate: `int_(-1)^1 (1 + x^2)/(9 - x^2) "d"x`
Evaluate: `int_0^(pi/4) log(1 + tanx) "d"x`
`int_0^(π/2) sin^6x cos^2x.dx` = ______.
Evaluate:
`int_(-π/2)^(π/2) |sinx|dx`
Evaluate:
`int_-4^5 |x + 3|dx`
Evaluate:
`int_0^(π/2) sinx/(1 + cosx)^3 dx`
Prove that: `int_0^1 logx/sqrt(1 - x^2)dx = π/2 log(1/2)`
If `int_0^π f(sinx)dx = kint_0^π f(sinx)dx`, then find the value of k.