Advertisements
Advertisements
Question
Prove that `int_0^"a" "f" ("x") "dx" = int_0^"a" "f" ("a" - "x") "d x",` hence evaluate `int_0^pi ("x" sin "x")/(1 + cos^2 "x") "dx"`
Advertisements
Solution
To prove: `int_0^a "f"("x") "dx" = int_0^a "f" ("a - x") "dx"`
Proof: Let t = a - x
⇒ dt = - dx
When x = 0, t = a
When x = a , t = 0
Putting the value of x in LHS
`int_a^0 "f"("a - t") (- "dt")`
= `- int_a^0 "f" ("a - t") ("dt")`
= `int_0^a "f" ("a - t") ("dt")`
= `int_0^a ("a - x") ("dx") ...(∵ int_a^b "f" (t) "dt" = int_a^b ("x")( "dx"))`
= RHS
Using this we can solve the given question as follows:
`I = int_0^pi f ("x") d"x" = int_0^pi (pi - "x") "dx"`
⇒ `2I = int_0^pi f ("x") d"x" + int_0^pi f (pi - "x") d"x" = int_0^pi ("x" sin "x")/(1 + cos^2 "x") d"x" + int_0^pi ((pi - "x") sin(pi - "x"))/(1 + cos^2 (pi - "x")) d"x"`
⇒`2"I" = int_0^pi ("x" sin "x")/(1 + cos^2 "x") "dx" + int_0^pi ((pi - "x")sin"x")/(1 + cos^2 (pi - "x")) "dx"`
⇒ `2"I" = int_0^pi (pi sin"x")/(1 + cos^2 "x") "dx"`
Let, cos x = t ⇒ -sin x dx = dt
⇒ `2"I" = -int_1^-1 (pi)/(1 + t^2) dt = -pi [ tan^-1 t ]_1^(-1) = -pi(-pi/(4) - pi/(4)) = pi^2/(2)`
∴ `"I" = pi^2/(4)`
RELATED QUESTIONS
By using the properties of the definite integral, evaluate the integral:
`int_(-5)^5 | x + 2| dx`
By using the properties of the definite integral, evaluate the integral:
`int_0^1 x(1-x)^n dx`
By using the properties of the definite integral, evaluate the integral:
`int_(pi/2)^(pi/2) sin^7 x dx`
By using the properties of the definite integral, evaluate the integral:
`int_0^(2x) cos^5 xdx`
By using the properties of the definite integral, evaluate the integral:
`int_0^pi log(1+ cos x) dx`
Evaluate : `int _0^(pi/2) "sin"^ 2 "x" "dx"`
Evaluate = `int (tan x)/(sec x + tan x)` . dx
`int_2^4 x/(x^2 + 1) "d"x` = ______
Evaluate `int_1^2 (sqrt(x))/(sqrt(3 - x) + sqrt(x)) "d"x`
By completing the following activity, Evaluate `int_2^5 (sqrt(x))/(sqrt(x) + sqrt(7 - x)) "d"x`.
Solution: Let I = `int_2^5 (sqrt(x))/(sqrt(x) + sqrt(7 - x)) "d"x` ......(i)
Using the property, `int_"a"^"b" "f"(x) "d"x = int_"a"^"b" "f"("a" + "b" - x) "d"x`, we get
I = `int_2^5 ("( )")/(sqrt(7 - x) + "( )") "d"x` ......(ii)
Adding equations (i) and (ii), we get
2I = `int_2^5 (sqrt(x))/(sqrt(x) - sqrt(7 - x)) "d"x + ( ) "d"x`
2I = `int_2^5 (("( )" + "( )")/("( )" + "( )")) "d"x`
2I = `square`
∴ I = `square`
`int_0^(pi/4) (sec^2 x)/((1 + tan x)(2 + tan x))`dx = ?
`int_0^1 (1 - x/(1!) + x^2/(2!) - x^3/(3!) + ... "upto" ∞)` e2x dx = ?
The value of `int_1^3 dx/(x(1 + x^2))` is ______
`int_0^pi sin^2x.cos^2x dx` = ______
`int_(-1)^1 (x^3 + |x| + 1)/(x^2 + 2|x| + 1) "d"x` is equal to ______.
`int_(-2)^2 |x cos pix| "d"x` is equal to ______.
`int_0^(2"a") "f"(x) "d"x = 2int_0^"a" "f"(x) "d"x`, if f(2a – x) = ______.
`int_0^(pi/2) sqrt(1 - sin2x) "d"x` is equal to ______.
Evaluate:
`int_2^8 (sqrt(10 - "x"))/(sqrt"x" + sqrt(10 - "x")) "dx"`
`int_0^1 1/(2x + 5) dx` = ______.
Let f be a real valued continuous function on [0, 1] and f(x) = `x + int_0^1 (x - t)f(t)dt`. Then, which of the following points (x, y) lies on the curve y = f(x)?
The value of `int_((-1)/sqrt(2))^(1/sqrt(2)) (((x + 1)/(x - 1))^2 + ((x - 1)/(x + 1))^2 - 2)^(1/2)`dx is ______.
`int_0^π(xsinx)/(1 + cos^2x)dx` equals ______.
If `β + 2int_0^1x^2e^(-x^2)dx = int_0^1e^(-x^2)dx`, then the value of β is ______.
Let `int_0^∞ (t^4dt)/(1 + t^2)^6 = (3π)/(64k)` then k is equal to ______.
Evaluate: `int_1^3 sqrt(x + 5)/(sqrt(x + 5) + sqrt(9 - x))dx`
Evaluate `int_-1^1 |x^4 - x|dx`.
Evaluate: `int_0^(π/4) log(1 + tanx)dx`.
Evaluate the following integral:
`int_0^1x(1-x)^5dx`
