Advertisements
Advertisements
प्रश्न
Fill in the blank : `int_4^9 (1)/sqrt(x)*dx` = _______
उत्तर
Let I = `int_4^9 (1)/sqrt(x)*dx`
= `int_4^9x^(1/2)*dx = [(x^(1/2))/(1/2)]_4^9`
= `2[sqrt(x)]_4^9`
= `2(sqrt(9) - sqrt(4))`
= 2 (3 – 2)
∴ I = 2.
APPEARS IN
संबंधित प्रश्न
Choose the correct option from the given alternatives :
`int_0^(pi/2) sn^6x cos^2x*dx` =
Evaluate the following : `int_(pi/4)^(pi/2) (cos theta)/[cos theta/2 + sin theta/2]^3*d theta`
Evaluate the following : `int_0^(pi/4) (tan^3x)/(1 +cos2x)*dx`
Evaluate the following : `int_0^pi x*sinx*cos^4x*dx`
Evaluate the following : If f(x) = a + bx + cx2, show that `int_0^1 f(x)*dx = (1/(6)[f(0) + 4f(1/2) + f(1)]`
Evaluate the following definite integral:
`int_(-2)^3 (1)/(x + 5)*dx`
Evaluate the following definite integrals: `int_1^2 dx/(x^2 + 6x + 5)`
Solve the following : `int_0^1 (1)/(sqrt(1 + x) + sqrt(x))dx`
`int_0^1 sqrt((1 - x)/(1 + x)) "d"x` =
Evaluate `int_0^1 (x^2 + 3x + 2)/sqrt(x) "d"x`
By completing the following activity, Evaluate `int_1^2 (x + 3)/(x(x + 2)) "d"x`
Solution: Let I = `int_1^2 (x + 3)/(x(x + 2)) "d"x`
Let `(x + 3)/(x(x + 2)) = "A"/x + "B"/((x + 2))`
∴ x + 3 = A(x + 2) + B.x
∴ A = `square`, B = `square`
∴ I = `int_1^2[("( )")/x + ("( )")/((x + 2))] "d"x`
∴ I = `[square log x + square log(x + 2)]_1^2`
∴ I = `square`
`int_(-5)^5 log ((7 - x)/(7 + x))`dx = ?
`int_0^1 tan^-1 ((2x - 1)/(1 + x - x^2))` dx = ?
Evaluate the following definite integral :
`int_1^2 (3"x")/((9"x"^2 - 1)) "dx"`
Evaluate:
`int_0^1 |x| dx`
Evaluate the following definite intergral:
`int_1^2 (3x)/((9x^2-1 )`dx
Evaluate the following definite intergral:
`int_-2^3 1/(x+5) · dx`
Evaluate the following definite integral:
`int_1^2 (3x)/((9x^2 - 1))*dx`
Evaluate the following definite intergral:
`\underset{4}{\overset{9}{int}}1/sqrt(x)dx`
