Advertisements
Advertisements
प्रश्न
Evaluate `int_0^1 (x^2 + 3x + 2)/sqrt(x) "d"x`
उत्तर
Let I = `int_0^1 (x^2 + 3x + 2)/sqrt(x) "d"x`
= `int_0^1 ((x^2 + 3x + 2)/x^(1/2)) "d"x`
= `int_0^1 (x^2/x^(1/2) + (3x)/(x^(1/2)) + 2/(x^(1/2))) "d"x`
= `int_0^1 (x^(3/2) + 3x^(1/2) + 2x^(-1/2)) "d"x`
= `int_0^1 x^(3/2) "d"x + 3int_0^1 x^(1/2) "d"x + 2int_0^1 x^(-1/2) "d"x`
= `[(x^(5/2))/(5/2)]_0^1 + 3[(x^(3/2))/(3/2)]_0^1 + 2[(x^(1/2))/(1/2)]_0^1`
= `2/5(1 - 0) + 3 xx 2/3(1 - 0) + 2 xx 2(1 - 0)`
= `2/5 + 2 + 4`
∴ I = `32/5`
APPEARS IN
संबंधित प्रश्न
Evaluate:
`int_(-pi/4)^(pi/4) (1)/(1 - sinx)*dx`
Evaluate : `int_0^(pi/4) (sec^2x)/(3tan^2x + 4tan x +1)*dx`
Evaluate : `int_(-1)^1 (1)/(a^2e^x + b^2e^(-x))*dx`
Choose the correct option from the given alternatives :
`int_0^(pi/2) sn^6x cos^2x*dx` =
Evaluate the following : `int_0^1 (1/(1 + x^2))sin^-1((2x)/(1 + x^2))*dx`
Evaluate the following definite integral:
`int_(-2)^3 (1)/(x + 5)*dx`
Choose the correct alternative :
`int_2^3 x/(x^2 - 1)*dx` =
Fill in the blank : `int_0^2 e^x*dx` = ________
Solve the following : `int_0^1 (x^2 + 3x + 2)/sqrt(x)*dx`
Solve the following : `int_3^5 dx/(sqrt(x + 4) + sqrt(x - 2)`
Prove that: `int_0^(2"a") "f"(x) "d"x = int_0^"a" "f"(x) "d"x + int_0^"a" "f"(2"a" - x) "d"x`
Evaluate the following definite integral:
`int_4^9 1/sqrtx dx`
Evaluate the following definite intergral:
`int_1^3 log xdx`
Evaluate the following definite integral:
`int_4^9 1/sqrt(x)dx`
Evaluate the following definite integral:
`int_1^3 logx dx`
Evaluate the following definite integral:
`int_-2^3 1/(x + 5) dx`
Evaluate the following definite intergral:
`int_1^3logxdx`
Solve the following.
`int_0 ^1 e^(x^2) * x^3`dx
Evaluate the following definite intergral:
`int_1^2(3x)/(9x^2-1).dx`
Solve the following.
`int_1^3x^2 logx dx`
