Advertisements
Advertisements
प्रश्न
Choose the correct alternative :
`int_2^7 sqrt(x)/(sqrt(x) + sqrt(9 - x))*dx` =
विकल्प
`(7)/(2)`
`(5)/(2)`
7
2
उत्तर
Let I = `int_2^7 sqrt(x)/(sqrt(x) + sqrt(9 - x))*dx` ...(i)
= `int_2^7 sqrt(2 + 7 - x)/(sqrt(2 + 7 - x) + sqrt(9 - (2 + 7 - x)))*dx ...[because int_"a"^"b" f(x)*dx = int_"a"^"b" f("a" + "b" - x)*dx]`
∴ I = `int_2^7 sqrt(9 - x)/(sqrt(9 - x) + sqrt(x))*dx` ...(ii)
Adding (i) and (ii), we get
2I = `int_2^7 sqrt(x)/(sqrt(x) + sqrt(9 - x))*dx + int_2^7 sqrt(9 - x)/(sqrt(9 - x) + sqrt(x))*dx`
= `int_2^7 (sqrt(x) + sqrt(9 - x))/(sqrt(x) + sqrt(9 - x))*dx`
= `int_2^7 1*dx`
= `[x]_2^7`
∴ 2I = 7 – 2 = 5
∴ I = `(5)/(2)`.
APPEARS IN
संबंधित प्रश्न
Evaluate : `int_(-1)^1 (1)/(a^2e^x + b^2e^(-x))*dx`
Evaluate the following : `int_(pi/4)^(pi/2) (cos theta)/[cos theta/2 + sin theta/2]^3*d theta`
Evaluate the following : `int_0^1 (cos^-1 x^2)*dx`
Evaluate the following definite integrals: `int_2^3 x/(x^2 - 1)*dx`
Choose the correct alternative :
`int_0^2 e^x*dx` =
Solve the following : `int_1^2 e^(2x) (1/x - 1/(2x^2))*dx`
Prove that: `int_"a"^"b" "f"(x) "d"x = int_"a"^"b" "f"("a" + "b" - x) "d"x`
Evaluate `int_1^"e" 1/(x(1 + log x)^2) "d"x`
Evaluate `int_0^"a" x^2 ("a" - x)^(3/2) "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`
Evaluate the following definite intergral:
`int_1^3 logx dx`
Evaluate the following definite intergral:
`int_-2^3 1/(x+5) dx`
`int_0^(π/2) (sin^2 x.dx)/(1 + cosx)^2` = ______.
`int_0^4 1/sqrt(4x - x^2)dx` = ______.
Solve the following.
`int_1^3 x^2 log x dx`
Solve the following.
`int_1^3 x^2 log x dx `
Evaluate the following definite intergral:
`int_(1)^3logx dx`
