हिंदी

If ∫abx3dx = 0, then (x4□)ab = 0 ⇒ 14(□-□) = 0 ⇒ b4 – □ = 0 ⇒ (b2 – a2)(□ + □) = 0 ⇒ b2 – □ = 0 as a2 + b2 ≠ 0 ⇒ b = ± □ - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

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`

रिक्त स्थान भरें
योग

उत्तर

`int_a^b x^3 dx` = 0, then `(x^4/bb4)_a^b` = 0

⇒ `1/4` (b4a4) = 0

⇒ b4a4 = 0

⇒ (b2 – a2)(b2 + a2) = 0

⇒ b2a2 = 0 as a2 + b2 ≠ 0

⇒ b = ± a

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2024-2025 (March) Model set 1 by shaalaa.com

संबंधित प्रश्न

Prove that: `int_0^(2a)f(x)dx=int_0^af(x)dx+int_0^af(2a-x)dx`


 
 

Evaluate : `intlogx/(1+logx)^2dx`

 
 

Evaluate : `intsec^nxtanxdx`


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

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


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_((-pi)/2)^(pi/2) sin^2 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 ______.


Evaluate `int e^x [(cosx - sin x)/sin^2 x]dx`


Evaluate`int (1)/(x(3+log x))dx` 


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


Using properties of definite integrals, evaluate 

`int_0^(π/2)  sqrt(sin x )/ (sqrtsin x + sqrtcos x)dx`


Evaluate the following integral:

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


State whether the following statement is True or False:

`int_(-5)^5 x/(x^2 + 7)  "d"x` = 10


`int_0^(pi/4) (sec^2 x)/((1 + tan x)(2 + tan x))`dx = ?


`int_0^1 (1 - x/(1!) + x^2/(2!) - x^3/(3!) + ... "upto" ∞)` e2x dx = ?


The value of `int_-3^3 ("a"x^5 + "b"x^3 + "c"x + "k")"dx"`, where a, b, c, k are constants, depends only on ______.


`int_0^{pi/2} log(tanx)dx` = ______


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


`int_2^3 x/(x^2 - 1)` dx = ______


`int_0^1 (1 - x)^5`dx = ______.


`int_0^{pi/2} cos^2x  dx` = ______ 


The value of `int_1^3 dx/(x(1 + x^2))` is ______ 


`int_0^{1/sqrt2} (sin^-1x)/(1 - x^2)^{3/2} dx` = ______ 


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


`int_(-1)^1 log ((2 - x)/(2 + x)) "dx" = ?`


`int_(pi/4)^(pi/2) sqrt(1-sin 2x)  dx =` ______.


`int_-1^1x^2/(1+x^2)  dx=` ______.


`int_(-1)^1 (x + x^3)/(9 - x^2)  "d"x` = ______.


Evaluate `int_(-1)^2 "f"(x)  "d"x`, where f(x) = |x + 1| + |x| + |x – 1|


If `int_0^1 "e"^"t"/(1 + "t") "dt"` = a, then `int_0^1 "e"^"t"/(1 + "t")^2 "dt"` is equal to ______.


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_((-pi)/4)^(pi/4) "dx"/(1 + cos2x)` is equal to ______.


`int_0^(pi/2)  cos x "e"^(sinx)  "d"x` is equal to ______.


`int_(-5)^5  x^7/(x^4 + 10)  dx` = ______.


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


Evaluate: `int_2^5 sqrt(x)/(sqrt(x) + sqrt(7) - x)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 ______.


`int_0^1|3x - 1|dx` equals ______.


`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 ______.


Evaluate: `int_0^π 1/(5 + 4 cos x)dx`


If `int_0^K dx/(2 + 18x^2) = π/24`, then the value of K is ______.


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


Evaluate `int_0^(π//4) log (1 + tanx)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`.


Evaluate: `int_(-π//4)^(π//4) (cos 2x)/(1 + cos 2x)dx`.


Evaluate: `int_0^π x/(1 + sinx)dx`.


Evaluate the following definite integral:

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


Evaluate:

`int_0^1 |2x + 1|dx`


Evaluate the following integral:

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


Evaluate the following definite intergral:

`int_1^2 (3x)/(9x^2 - 1) dx`


Evaluate the following integral:

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×