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 : `intsec^nxtanxdx`


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

`int_0^(pi/2) (2log sin x - log sin 2x)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.


Evaluate `int_0^(pi/2) cos^2x/(1+ sinx cosx) dx`


Evaluate : `int 1/("x" [("log x")^2 + 4])  "dx"`


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


Evaluate the following integrals : `int_2^5 sqrt(x)/(sqrt(x) + sqrt(7 - x))*dx`


`int_"a"^"b" "f"(x)  "d"x` = ______


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


`int_0^1 log(1/x - 1) "dx"` = ______.


`int_0^{pi/2} (cos2x)/(cosx + sinx)dx` = ______


`int_0^9 1/(1 + sqrtx)` dx = ______ 


Evaluate the following:

`int_0^(pi/2)  "dx"/(("a"^2 cos^2x + "b"^2 sin^2 x)^2` (Hint: Divide Numerator and Denominator by cos4x)


`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:


`int_4^9 1/sqrt(x)dx` = ______.


`int_0^5 cos(π(x - [x/2]))dx` where [t] denotes greatest integer less than or equal to t, is equal to ______.


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.


The integral `int_0^2||x - 1| -x|dx` is equal to ______.


`int_0^π(xsinx)/(1 + cos^2x)dx` equals ______.


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 ______.


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


The value of the integral `int_0^1 x cot^-1(1 - x^2 + x^4)dx` is ______.


What is `int_0^(π/2)` sin 2x ℓ n (cot x) dx equal to ?


`int_(π/3)^(π/2) x sin(π[x] - x)dx` is equal to ______.


`int_0^(π/4) x. sec^2 x  dx` = ______.


`int_((-π)/2)^(π/2) log((2 - sinx)/(2 + sinx))` is equal to ______.


Evaluate: `int_1^3 sqrt(x + 5)/(sqrt(x + 5) + sqrt(9 - x))dx`


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`.


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


Solve the following.

`int_1^3 x^2 logx  dx`


`int_1^2 x logx  dx`= ______


Evaluate the following definite integral:

`int_1^3 log x  dx`


Evaluate the following definite integral:

`int_-2^3 1/(x + 5) dx`


Evaluate the following integral:

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


Evaluate:

`int_0^sqrt(2)[x^2]dx`


`∫_0^(π/2) (sqrttan x + sqrtcot x)dx` = ______.


`int_0^(pi/4) (cos^2 x)/(cos^2 x + 4 sin^2 x) dx` =


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×