Advertisements
Advertisements
प्रश्न
By using the properties of the definite integral, evaluate the integral:
`int_0^a sqrtx/(sqrtx + sqrt(a-x)) dx`
उत्तर
Let I = `int_0^a (sqrtx)/(sqrtx + sqrt(a - x)) dx` ....(i)
`= I = int_0^a (sqrt(a - x))/(sqrt(a - x) + sqrt (a - (a - x)))`
I = `int_0^a sqrt(a - x)/(sqrt(a - x) + sqrtx) dx` ....(ii)
`[because int_0^a f(x) dx = int_0^a f(a - x) dx]`
On adding equation (i) and (ii),
2 I = `int_0^a (sqrtx + sqrt(a - x))/(sqrt(a - x) + sqrtx) dx`
2 I `= int_0^a 1 * dx => [x]_0^a`
⇒ 2I = a
∴ `I = a/2`
APPEARS IN
संबंधित प्रश्न
Evaluate : `int e^x[(sqrt(1-x^2)sin^-1x+1)/(sqrt(1-x^2))]dx`
Evaluate : `intsec^nxtanxdx`
By using the properties of the definite integral, evaluate the integral:
`int_0^(pi/2) (cos^5 xdx)/(sin^5 x + cos^5 x)`
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.
Evaluate the definite integrals `int_0^pi (x tan x)/(sec x + tan x)dx`
Prove that `int_0^af(x)dx=int_0^af(a-x) dx`
hence evaluate `int_0^(pi/2)sinx/(sinx+cosx) dx`
Evaluate : `int "e"^(3"x")/("e"^(3"x") + 1)` dx
Using properties of definite integrals, evaluate
`int_0^(π/2) sqrt(sin x )/ (sqrtsin x + sqrtcos x)dx`
`int_1^2 1/(2x + 3) dx` = ______
Evaluate `int_1^2 (sqrt(x))/(sqrt(3 - x) + sqrt(x)) "d"x`
`int_0^1 ((x^2 - 2)/(x^2 + 1))`dx = ?
`int_-9^9 x^3/(4 - x^2)` dx = ______
If `int_0^"a" sqrt("a - x"/x) "dx" = "K"/2`, then K = ______.
`int_(pi/18)^((4pi)/9) (2 sqrt(sin x))/(sqrt (sin x) + sqrt(cos x))` dx = ?
`int_0^1 x tan^-1x dx` = ______
If `int_0^"k" "dx"/(2 + 32x^2) = pi/32,` then the value of k is ______.
The value of `int_2^7 (sqrtx)/(sqrt(9 - x) + sqrtx)dx` is ______
`int_0^(pi/2) 1/(1 + cos^3x) "d"x` = ______.
Evaluate `int_(-1)^2 "f"(x) "d"x`, where f(x) = |x + 1| + |x| + |x – 1|
`int_("a" + "c")^("b" + "c") "f"(x) "d"x` is equal to ______.
If `int_0^1 "e"^"t"/(1 + "t") "dt"` = a, then `int_0^1 "e"^"t"/(1 + "t")^2 "dt"` is equal to ______.
`int_(-2)^2 |x cos pix| "d"x` is equal to ______.
`int_0^(pi/2) (sin^"n" x"d"x)/(sin^"n" x + cos^"n" x)` = ______.
Evaluate:
`int_2^8 (sqrt(10 - "x"))/(sqrt"x" + sqrt(10 - "x")) "dx"`
Evaluate: `int_2^5 sqrt(x)/(sqrt(x) + sqrt(7) - x)dx`
`int_a^b f(x)dx` = ______.
If `int_0^1(sqrt(2x) - sqrt(2x - x^2))dx = int_0^1(1 - sqrt(1 - y^2) - y^2/2)dy + int_1^2(2 - y^2/2)dy` + I then I equal.
Let a be a positive real number such that `int_0^ae^(x-[x])dx` = 10e – 9 where [x] is the greatest integer less than or equal to x. Then, a is equal to ______.
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 ______.
`int_0^(pi/4) (sec^2x)/((1 + tanx)(2 + tanx))dx` equals ______.
Evaluate: `int_0^π 1/(5 + 4 cos x)dx`
With the usual notation `int_1^2 ([x^2] - [x]^2)dx` is equal to ______.
If `int_0^(π/2) log cos x dx = π/2 log(1/2)`, then `int_0^(π/2) log sec dx` = ______.
`int_((-π)/2)^(π/2) log((2 - sinx)/(2 + sinx))` is equal to ______.
`int_0^(π/2)((root(n)(secx))/(root(n)(secx + root(n)("cosec" x))))dx` is equal to ______.
Evaluate: `int_0^π x/(1 + sinx)dx`.
For any integer n, the value of `int_-π^π e^(cos^2x) sin^3 (2n + 1)x dx` is ______.
Evaluate : `int_-1^1 log ((2 - x)/(2 + x))dx`.
Evaluate the following integral:
`int_-9^9 x^3/(4 - x^2) dx`
Solve the following.
`int_0^1e^(x^2)x^3 dx`
