मराठी

By using the properties of the definite integral, evaluate the integral: ∫0a xx +a-x dx

Advertisements
Advertisements

प्रश्न

By using the properties of the definite integral, evaluate the integral:

`int_0^a  sqrtx/(sqrtx + sqrt(a-x))   dx`

बेरीज
Advertisements

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`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Integrals - Exercise 7.11 [पृष्ठ ३४७]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
पाठ 7 Integrals
Exercise 7.11 | Q 17 | पृष्ठ ३४७

संबंधित प्रश्‍न

Evaluate : `int e^x[(sqrt(1-x^2)sin^-1x+1)/(sqrt(1-x^2))]dx`


 
 

Evaluate : `intlogx/(1+logx)^2dx`

 
 

By using the properties of the definite integral, evaluate the integral:

`int_0^(pi/4) log (1+ tan x) dx`


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.


`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`int (1)/(x(3+log x))dx` 


Evaluate : `int _0^(pi/2) "sin"^ 2  "x"  "dx"`


Prove that `int _a^b f(x) dx = int_a^b f (a + b -x ) dx`  and hence evaluate   `int_(pi/6)^(pi/3) (dx)/(1 + sqrt(tan x))` .   


Evaluate  : `int "x"^2/("x"^4 + 5"x"^2 + 6) "dx"`


Find `dy/dx, if y = cos^-1 ( sin 5x)`


Evaluate: `int_0^pi ("x"sin "x")/(1+ 3cos^2 "x") 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)` log(1 + tanθ) dθ = ______


`int_0^4 1/(1 + sqrtx)`dx = ______.


`int_-2^1 dx/(x^2 + 4x + 13)` = ______


`int_0^pi sin^2x.cos^2x  dx` = ______ 


The value of `int_2^7 (sqrtx)/(sqrt(9 - x) + sqrtx)dx` is ______ 


`int_(-1)^1 (x^3 + |x| + 1)/(x^2 + 2|x| + 1) "d"x` is equal to ______.


`int_0^(2"a") "f"(x) "d"x = 2int_0^"a" "f"(x) "d"x`, if f(2a – x) = ______.


If `f(a + b - x) = f(x)`, then `int_0^b x f(x)  dx` is equal to


Evaluate: `int_1^3 sqrt(x)/(sqrt(x) + sqrt(4) - x) dx`


`int_0^1 1/(2x + 5) dx` = ______.


`int_a^b f(x)dx = int_a^b f(x - a - b)dx`.


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 ______.


The value of the integral `int_0^sqrt(2)([sqrt(2 - x^2)] + 2x)dx` (where [.] denotes greatest integer function) is ______.


`int_-1^1 (17x^5 - x^4 + 29x^3 - 31x + 1)/(x^2 + 1) dx` is equal to ______.


If `int_0^(2π) cos^2 x  dx = k int_0^(π/2) cos^2 x  dx`, then the value of k is ______.


The value of `int_0^(π/4) (sin 2x)dx` is ______.


Evaluate the following integral:

`int_0^1 x(1 - 5)^5`dx


 `int_-9^9 x^3/(4-x^2) dx` =______


Evaluate: `int_-1^1 x^17.cos^4x  dx`


Evaluate the following integral:

`int_-9^9 x^3/(4 - x^2) dx`


Evaluate the following integral:

`int_-9^9 x^3/(4-x^2)dx`


`int_(pi"/"11)^(9pi"/"22) (dx)/(1 + sqrttan x)` =


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×