Advertisements
Advertisements
Question
Evaluate the following : `int_(-3)^(3) x^3/(9 - x^2)*dx`
Solution
Let I = `int_(-3)^(3) x^3/(9 - x^2)*dx`
Let f(x) = `x^3/(9 - x^2)`
∴ f( –x) = `(-x)^3/(9 - (- x)^2`
= `(-x^3)/(9 - x^2)`
= `-f(x)`
∴ f is an odd function.
∴ `int_-3^3 f(x)*dx = 0, "i.e." int_-3^3 x^3/(9 - x^2)*dx` = 0.
APPEARS IN
RELATED QUESTIONS
Evaluate : `int_3^5 (1)/(sqrt(2x + 3) - sqrt(2x - 3))*dx`
Evaluate : `int_0^(pi/4) sin^4x*dx`
Evaluate : `int_(-4)^2 (1)/(x^2 + 4x + 13)*dx`
Evaluate:
`int_0^(pi/2) sqrt(cos x) sin^3x * dx`
Evaluate : `int_1^3 (cos(logx))/x*dx`
Evaluate the following:
`int_((-pi)/2)^(pi/2) log((2 + sin x)/(2 - sin x)) * dx`
Choose the correct option from the given alternatives :
`int_1^2 (1)/x^2 e^(1/x)*dx` =
Choose the correct option from the given alternatives :
The value of `int_((-pi)/4)^(pi/4) log((2+ sin theta)/(2 - sin theta))*d theta` is
Evaluate the following : `int_0^(pi/2) cosx/(3cosx + sinx)*dx`
Evaluate the following : `int_0^1 (cos^-1 x^2)*dx`
Evaluate the following : `int_0^pi x*sinx*cos^4x*dx`
Evaluate the following : `int_0^a 1/(a^2 + ax - x^2)*dx`
Evaluate the following definite integrals: `int_0^1 (1)/(sqrt(1 + x) + sqrt(x))*dx`
Fill in the blank : `int_2^3 x/(x^2 - 1)*dx` = _______
State whether the following is True or False : `int_"a"^"b" f(x)*dx = int_"a"^"b" f("t")*dt`
State whether the following is True or False : `int_"a"^"b" f(x)*dx = int_"a"^"b" f(x - "a" - "b")*dx`
State whether the following is True or False : `int_1^2 sqrt(x)/(sqrt(3 - x) + sqrt(x))*dx = (1)/(2)`
Solve the following:
`int_0^1 e^(x^2)*x^3dx`
Solve the following : `int_3^5 dx/(sqrt(x + 4) + sqrt(x - 2)`
Solve the following : `int_0^9 (1)/(1 + sqrt(x))*dx`
Choose the correct alternative:
`int_4^9 ("d"x)/sqrt(x)` =
State whether the following statement is True or False:
`int_2^3 x/(x^2 + 1) "d"x = 1/2 log 2`
Evaluate `int_1^"e" 1/(x(1 + log x)^2) "d"x`
Evaluate:
`int_1^2 1/(x^2 + 6x + 5) dx`
Evaluate `int_2^3 x/((x + 2)(x + 3)) "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_0^(pi/2) root(7)(sin x)/(root(7)(sin x) + root(7)(cos x))`dx = ?
Prove that: `int_0^(2a) f(x)dx = int_0^a f(x)dx + int_0^a f(2a - x)dx`
Evaluate the following definite intergrals.
`int_1^3 logx* dx`
Evaluate the following definite integral :
`int_1^2 (3"x")/((9"x"^2 - 1)) "dx"`
Evaluate the following integrals:
`int_0^1 x(1 - x)^5 dx`
Evaluate:
`int_(-π/2)^(π/2) (sin^3x)/(1 + cos^2x)dx`
Evaluate the following definite intergral:
`int_4^9 1/sqrt(x)dx`
Solve the following.
`int_1^3 x^2 log x dx`
Solve the following.
`int_1^3 x^2 log x dx `
`int_a^b f(x) dx = int_a^b f (t) dt`
Evaluate the following definite intergral:
`int_1^3logxdx`
Evaluate the following definite intergral:
`int_(-2)^3 1/(x + 5)dx`
