Advertisements
Advertisements
प्रश्न
Prove that: `int_0^(2a)f(x)dx=int_0^af(x)dx+int_0^af(2a-x)dx`
उत्तर
`LHS=int_0^(2a)f(x)dx=int_0^af(x)dx+int_a^(2a)f(x)dx.........(1)`
Substitute x = a + t in the second integral
dx=dt
When x = a, t = 0.
When x = 2a, t = a.
`thereforeint_a^(2a)f(x)dx=int_0^af(a+t)dt`
`=int_0^af(a+(a-t))dt (therefore int_0^af(x)dx=int_0^a f(a-x)dx)`
`=int_0^af(2a-t)dt`
`int_a^(2a)f(x)dx=int_0^af(2a-x)dx (therefore int_0^af(t)dt=int_0^af(x)dx)`
Using the above in (1), we get
`int_0^(2a)f(x)dx=int_0^af(x)dx+int_a^(2a)f(x)dx`
`=int_0^af(x)dx+int_0^af(2a-x)dx=RHS ("Proved")`
APPEARS IN
संबंधित प्रश्न
Evaluate : `intsec^nxtanxdx`
By using the properties of the definite integral, evaluate the integral:
`int_0^(pi/4) log (1+ tan x) dx`
By using the properties of the definite integral, evaluate the integral:
`int_0^(2x) cos^5 xdx`
`int_(-pi/2)^(pi/2) (x^3 + x cos x + tan^5 x + 1) dx ` is ______.
The value of `int_0^(pi/2) log ((4+ 3sinx)/(4+3cosx))` dx is ______.
Evaluate the definite integrals `int_0^pi (x tan x)/(sec x + tan x)dx`
Evaluate `int_0^(pi/2) cos^2x/(1+ sinx cosx) dx`
\[\int\limits_0^a 3 x^2 dx = 8,\] find the value of a.
Evaluate : `int 1/("x" [("log x")^2 + 4]) "dx"`
Evaluate : `int 1/sqrt("x"^2 - 4"x" + 2) "dx"`
`int_2^7 sqrt(x)/(sqrt(x) + sqrt(9 - x)) dx` = ______.
`int_0^1 "e"^(2x) "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)` log(1 + tanθ) dθ = ______
`int_2^3 x/(x^2 - 1)` dx = ______
`int_0^{pi/2} cos^2x 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 =` ______.
`int_0^(pi/2) 1/(1 + cosx) "d"x` = ______.
`int_0^(pi/2) sqrt(1 - sin2x) "d"x` is equal to ______.
`int_0^(2"a") "f"("x") "dx" = int_0^"a" "f"("x") "dx" + int_0^"a" "f"("k" - "x") "dx"`, then the value of k is:
If f(x) = `(2 - xcosx)/(2 + xcosx)` and g(x) = logex, (x > 0) then the value of the integral `int_((-π)/4)^(π/4) "g"("f"(x))"d"x` is ______.
Evaluate: `int_0^π 1/(5 + 4 cos x)dx`
`int_0^(π/2)((root(n)(secx))/(root(n)(secx + root(n)("cosec" x))))dx` is equal to ______.
Assertion (A): `int_2^8 sqrt(10 - x)/(sqrt(x) + sqrt(10 - x))dx` = 3.
Reason (R): `int_a^b f(x) dx = int_a^b f(a + b - x) dx`.
Evaluate: `int_(-π//4)^(π//4) (cos 2x)/(1 + cos 2x)dx`.
Evaluate the following definite integral:
`int_-2^3 1/(x + 5) dx`
Evaluate: `int_-1^1 x^17.cos^4x dx`
Evaluate:
`int_0^1 |2x + 1|dx`
Evaluate the following integral:
`int_0^1 x(1 - x)^5 dx`
Evaluate the following integral:
`int_-9^9x^3/(4-x^2)dx`
Evaluate:
`int_0^6 |x + 3|dx`
Evaluate the following definite intergral:
`int_1^3logx dx`
