हिंदी

∫02exdx = ______. - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

`int_0^2 e^x dx` = ______.

विकल्प

  • e2 – 1

  • 1 – e2 

  • e – 1

  • 1 – e

MCQ
रिक्त स्थान भरें

उत्तर

`int_0^2 e^x dx` = `bb(underline(e^2 - 1))`.

Explanation:

`int_0^2 e^xdx = [e^x]_0^2`

= e2 – e0

= e2 – 1

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1.6: Definite Integration - Q.1

संबंधित प्रश्न

Evaluate :`int_0^pi(xsinx)/(1+sinx)dx`


 
 

Evaluate `int_(-2)^2x^2/(1+5^x)dx`

 
 

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

`int_0^(pi/2) cos^2 x dx`


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

`int_0^(pi/2)  (cos^5  xdx)/(sin^5 x + cos^5 x)`


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

`int_0^2 xsqrt(2 -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.


The value of `int_0^(pi/2) log  ((4+ 3sinx)/(4+3cosx))` dx is ______.


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


\[\int\limits_0^k \frac{1}{2 + 8 x^2} dx = \frac{\pi}{16},\] find the value of k.


If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that

\[\int_a^b xf\left( x \right)dx = \left( \frac{a + b}{2} \right) \int_a^b f\left( x \right)dx\]

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  "e"^(3"x")/("e"^(3"x") + 1)` dx


Evaluate :  ∫ log (1 + x2) dx


Using properties of definite integrals, evaluate 

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


Find : `int_  (2"x"+1)/(("x"^2+1)("x"^2+4))d"x"`.


Evaluate the following integral:

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


`int_(-7)^7 x^3/(x^2 + 7)  "d"x` = ______


State whether the following statement is True or False:

`int_(-5)^5 x/(x^2 + 7)  "d"x` = 10


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 = ______


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^{pi/2}((3sqrtsecx)/(3sqrtsecx + 3sqrt(cosecx)))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` = ______ 


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


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


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


Show that `int_0^(pi/2) (sin^2x)/(sinx + cosx) = 1/sqrt(2) log (sqrt(2) + 1)`


`int_("a" + "c")^("b" + "c") "f"(x) "d"x` 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)


Evaluate the following:

`int_(-pi/4)^(pi/4) log|sinx + cosx|"d"x`


`int_((-pi)/4)^(pi/4) "dx"/(1 + cos2x)` 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:


The value of `int_0^1 tan^-1 ((2x - 1)/(1 + x - x^2))  dx` is


Evaluate: `int_(pi/6)^(pi/3) (dx)/(1 + sqrt(tanx)`


Evaluate: `int_(-1)^3 |x^3 - x|dx`


Evaluate: `int_((-π)/2)^(π/2) (sin|x| + cos|x|)dx`


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`


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


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


`int_-1^1 |x - 2|/(x - 2) dx`, x ≠ 2 is equal to ______.


For any integer n, the value of `int_-π^π e^(cos^2x) sin^3 (2n + 1)x  dx` is ______.


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


Evaluate the following integral:

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


`int_1^2 x logx  dx`= ______


Evaluate the following definite integral:

`int_1^3 log x  dx`


Evaluate: `int_-1^1 x^17.cos^4x  dx`


Evaluate:

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


Evaluate the following integral:

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


Evaluate the following integral:

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×