Advertisements
Advertisements
प्रश्न
Show that `int_0^a f(x)g (x)dx = 2 int_0^a f(x) dx` if f and g are defined as f(x) = f(a-x) and g(x) + g(a-x) = 4.
Advertisements
उत्तर
Let `I = int_0^a f(x) g(x) dx`
`= int_0^a f(a - x) [4 - g(a - x)] dx`
`= 4 int_0^a f(a - x) dx - int_0^a f(a - x) g (a - x) dx`
Let a - x = t
⇒ - dx = dt
When x = 0, t = a
and x = a, t = 0
`I = -4 int_a^0 f (t) dt + int_a^0 f (t) g (t) dt`
`= 4 int_0^a f (t) dt - int_0^a f (t) g (t) dt`
`= 4 int_0^a f (x) dx - int_0^a f (x)g (x) dx `
`= 4 int_0^a f (x) dx - I`
⇒ `2I = 4 int_0^a f (x) dx`
Hence, `I = 2 int_0^a f (x) dx`
APPEARS IN
संबंधित प्रश्न
If `int_0^alpha3x^2dx=8` then the value of α is :
(a) 0
(b) -2
(c) 2
(d) ±2
By using the properties of the definite integral, evaluate the integral:
`int_((-pi)/2)^(pi/2) sin^2 x dx`
By using the properties of the definite integral, evaluate the integral:
`int_0^a sqrtx/(sqrtx + sqrt(a-x)) dx`
`int_(-pi/2)^(pi/2) (x^3 + x cos x + tan^5 x + 1) dx ` is ______.
\[\int\limits_0^a 3 x^2 dx = 8,\] find the value of a.
Evaluate : \[\int(3x - 2) \sqrt{x^2 + x + 1}dx\] .
Evaluate : `int _0^(pi/2) "sin"^ 2 "x" "dx"`
Evaluate the following integral:
`int_0^1 x(1 - x)^5 *dx`
`int_0^1 "e"^(2x) "d"x` = ______
`int_0^1 (1 - x/(1!) + x^2/(2!) - x^3/(3!) + ... "upto" ∞)` e2x dx = ?
`int_0^{pi/2} log(tanx)dx` = ______
`int_"a"^"b" sqrtx/(sqrtx + sqrt("a" + "b" - x)) "dx"` = ______.
If `int_0^"a" sqrt("a - x"/x) "dx" = "K"/2`, then K = ______.
`int_3^9 x^3/((12 - x)^3 + x^3)` dx = ______
`int_0^{1/sqrt2} (sin^-1x)/(1 - x^2)^{3/2} dx` = ______
`int_0^pi sin^2x.cos^2x dx` = ______
`int_0^1 log(1/x - 1) "dx"` = ______.
`int_(pi/4)^(pi/2) sqrt(1-sin 2x) dx =` ______.
Which of the following is true?
Find `int_2^8 sqrt(10 - x)/(sqrt(x) + sqrt(10 - x)) "d"x`
Evaluate the following:
`int_(-pi/4)^(pi/4) log|sinx + cosx|"d"x`
`int_(-5)^5 x^7/(x^4 + 10) dx` = ______.
If `int_a^b x^3 dx` = 0, then `(x^4/square)_a^b` = 0
⇒ `1/4 (square - square)` = 0
⇒ b4 – `square` = 0
⇒ (b2 – a2)(`square` + `square`) = 0
⇒ b2 – `square` = 0 as a2 + b2 ≠ 0
⇒ b = ± `square`
`int_4^9 1/sqrt(x)dx` = ______.
Let `int ((x^6 - 4)dx)/((x^6 + 2)^(1/4).x^4) = (ℓ(x^6 + 2)^m)/x^n + C`, then `n/(ℓm)` is equal to ______.
Let f be continuous periodic function with period 3, such that `int_0^3f(x)dx` = 1. Then the value of `int_-4^8f(2x)dx` is ______.
If `int_0^K dx/(2 + 18x^2) = π/24`, then the value of K is ______.
For any integer n, the value of `int_-π^π e^(cos^2x) sin^3 (2n + 1)x dx` is ______.
If `int_0^1(3x^2 + 2x+a)dx = 0,` then a = ______
`int_1^2 x logx dx`= ______
Solve.
`int_0^1e^(x^2)x^3dx`
Evaluate:
`int_0^sqrt(2)[x^2]dx`
Evaluate the following integral:
`int_0^1x(1 - x)^5dx`
`int_(pi"/"11)^(9pi"/"22) (dx)/(1 + sqrttan x)` =
Which formula is \[P_5\] : Halving the Upper Limit (Addition)?
If \[f(-x)=-f(x)\], what is \[\int_{-a}^{a} f(x)\,dx\]?
If \[I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{dx}{1+\sqrt{\tan x}}\], what is the value of \[I\]?
