Advertisements
Advertisements
प्रश्न
Choose the correct alternative :
`int_(-9)^9 x^3/(4 - x^2)*dx` =
विकल्प
0
3
9
– 9
उत्तर
Let I = `int_(-9)^9 x^3/(4 - x^2)*dx`
Let f(x) = `x^3/(4 - x^2)`
∴ f(– x) = `(-x)^2/(4 - (-x)^2`
= `-x^3/(4 - x^2)`
= – f(x)
∴ f(x) is an odd function.
∴ `int_(-9)^9 x^3/(4 - x^2)*dx = 0. ...[because int_("a")^"a" f(x) = 0, if f(x) "odd function"]`
APPEARS IN
संबंधित प्रश्न
Evaluate : `int_3^5 (1)/(sqrt(2x + 3) - sqrt(2x - 3))*dx`
Evaluate : `int_0^(pi//4) (sin2x)/(sin^4x + cos^4x)*dx`
Evaluate : `int_0^(pi/4) sec^4x*dx`
Evaluate the following:
`int_((-pi)/2)^(pi/2) log((2 + sin x)/(2 - sin x)) * dx`
Evaluate the following : `int_0^pi x sin x cos^2x*dx`
Evaluate the following : `int_0^(pi/2) cosx/(3cosx + sinx)*dx`
Evaluate the following definite integrals: if `int_1^"a" (3x^2 + 2x + 1)*dx` = 11, find a.
Choose the correct alternative :
`int_2^3 x/(x^2 - 1)*dx` =
`int_1^9 (x + 1)/sqrt(x) "d"x` =
Choose the correct alternative:
`int_2^3 x^4 "d"x` =
Evaluate `int_0^"a" x^2 ("a" - x)^(3/2) "d"x`
Evaluate the following definite integrats:
`int_4^9 1/sqrt x dx`
Evaluate the following definite intergral:
`int_1^3 logx dx`
Evaluate the following definite integral:
`int_1^3 log x dx`
Evaluate the following definite integral:
`int_4^9 1/sqrtx dx`
Solve the following.
`int_0 ^1 e^(x^2) * x^3`dx
Evaluate the following definite intergral:
`int_4^9 1/sqrtx dx`
Evaluate the following definite intergral:
`int_1^3 log x dx`
Evaluate the following definite intergral:
`int_1^2(3x)/((9x^2 - 1))dx`
