Advertisements
Advertisements
प्रश्न
Solve the following : `int_0^1 (1)/(2x - 3)*dx`
उत्तर
Let I = `int_0^1 (1)/(2x - 3)*dx`
Put 2x – 3 =t
∴ 2·dx = dt
∴ dx = `"dt"/(2)`
When x = 0t = 2(0) – 3 = – 3
When x = 1, t = 2(1) – 3 = – 1
∴ I = `int_(-3)^(-1) (1)/"t"*"dt"/(2)`
= `(1)/(2) int_(-3)^(-1) "dt"/"t"`
= `(1)/(2)[log |"t"|]_(-3)^(-1)`
= `(1)/(2)[log|-1| - log|-3|]`
= `(1)/(2)(log 1 - log 3)`
= `(1)/(2)(0 - log 3)`
∴ I = `-(1)/(2) log 3`.
APPEARS IN
संबंधित प्रश्न
Prove that:
`int 1/(a^2 - x^2) dx = 1/2 a log ((a +x)/(a-x)) + c`
Evaluate : `int_0^4 (1)/sqrt(4x - x^2)*dx`
Evaluate: `int_0^(pi/2) sin2x*tan^-1 (sinx)*dx`
Evaluate the following : `int_0^(pi/4) (tan^3x)/(1 +cos2x)*dx`
Evaluate the following definite integral:
`int_1^3 logx.dx`
Fill in the blank : `int_4^9 (1)/sqrt(x)*dx` = _______
Solve the following : `int_3^5 dx/(sqrt(x + 4) + sqrt(x - 2)`
Solve the following : `int_0^4 (1)/sqrt(x^2 + 2x + 3)*dx`
Choose the correct alternative:
`int_(-2)^3 1/(x + 5) "d"x` =
Evaluate `int_1^3 log x "d"x`
Solve the following `int_1^3 x^2log x dx`
`int_0^(π/2) (sin^2 x.dx)/(1 + cosx)^2` = ______.
Evaluate the following definite integral:
`int_-2^3 1/(x+5) *dx`
The principle solutions of the equation cos θ = `1/2` are ______.
Evaluate the following definite intergral:
`int_1^2(3x)/((9x^2-1))dx`
Evaluate the following definite intergral:
`int_4^9(1)/sqrtxdx`
Evaluate the following definite intergral:
`int_-2^3 1/(x+5) · dx`
Evaluate the following definite intergral:
`int_-2^3 1/(x+5)dx`
Evaluate the following definite intergral:
`int_(-2)^3 1/(x + 5)dx`
