English

Evaluate the following as limit of sum: d∫02(x2+3)dx - Mathematics

Advertisements
Advertisements

Question

Evaluate the following as limit of sum:

`int _0^2 (x^2 + 3) "d"x`

Sum
Advertisements

Solution

We know that `int_"a"^"b" "f"(x) "d"x = lim_("n" -> oo) "h" sum_("r" = 0)^("n" - 1) "f"("a" + "rh")`

For I = `int_0^2 (x^2 + 3) "d"x`

We have a = 0 and b = 2

I = `int_00^2 (x^2 + 3) "d"x`

Here, a = 0, b = 2 and h = `("b" - "a")/"n" = (2 - 0)/"n" = 2/"n"`

⇒ nh = 2

And f(x) = `(x^2 + 3)`

∴ I = `int_0^2 (x^2 + 3)"d"x = lim_("h" -> 0) "h" sum_("r" = 0)^("n" - 1) "f"("a" + "rh")`

= `lim_("h" -> 0) "h" sum_("r" = 0)^("n" - 1) "f"("rh")`

= `lim_("h" -> 0) "h" sum_("r" = 0)^("n" - 1) (3 + "r"^2"h"^2)`

= `lim_("h" -> 0) "h"[3"n" + "h"^2 ((("n" - 1)("n" - 1 + 1)(2"n" - 2 + 1))/6)]`

= `lim_("h" -> 0) "h"[3"n" + "h"^2 ((("n"^2 - "n")(2"n" - 1))/6)]`

= `lim_("h" -> 0) "h" [3"n" + "h"^2/6 (2"n"^3 - 3"n"^2 + "n")]`

= `lim_("h" -> 0) [3"nh" + (2"n"^3"h"^3 - 3"n"^2"h"^2 * "h" + "nh" * "h"^2)/6]`

= `lim_("h" -> 0) [3.2 + (2.2^3 - 3.2^2 * "h" + 2 * "h"^2)/6]`

= `6 + 16/6`

= `26/3`

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Integrals - Exercise [Page 165]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 12
Chapter 7 Integrals
Exercise | Q 27 | Page 165

RELATED QUESTIONS

Evaluate the following definite integrals as limit of sums.

`int_1^4 (x^2 - x) dx`


Evaluate the definite integral:

`int_(pi/2)^pi e^x ((1-sinx)/(1-cos x)) dx`


Evaluate the definite integral:

`int_(pi/6)^(pi/3)  (sin x + cosx)/sqrt(sin 2x) dx`


Prove the following:

`int_1^3 dx/(x^2(x +1)) = 2/3 + log  2/3`


Prove the following:

`int_0^1 xe^x dx = 1`


Prove the following:

`int_0^(pi/4) 2 tan^3 xdx = 1 - log 2`


Evaluate  `int_0^1 e^(2-3x) dx` as a limit of a sum.


`int dx/(e^x + e^(-x))` is equal to ______.


`int (cos 2x)/(sin x + cos x)^2dx` is equal to ______.


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

(A) 1

(B) 0

(C) –1

(D) `pi/4`


Evaluate : `int_1^3 (x^2 + 3x + e^x) dx` as the limit of the sum.


` ∫  log x / x  dx `
 
 
 

\[\int\frac{\sin^3 x}{\sqrt{\cos x}} dx\]

\[\int\frac{1}{\sqrt{\tan^{- 1} x} . \left( 1 + x^2 \right)} dx\]

\[\int e^{cos^2 x}   \text{sin 2x  dx}\]

\[\int\frac{\log x^2}{x} dx\]

\[\int\cot x \cdot \log \text{sin x dx}\]

\[\text{ ∫  cosec x  log}      \left( \text{cosec x} - \cot x \right) dx\]

\[\int\frac{1}{x^2} \cos^2 \left( \frac{1}{x} \right) dx\]

\[\int4 x^3 \sqrt{5 - x^2} dx\]

\[\int\limits_0^1 \left( x e^x + \cos\frac{\pi x}{4} \right) dx\]

 


\[\int\limits_{- \pi/2}^{\pi/2} \sin^4 x\ dx\]

Evaluate the following integrals as limit of sums:

\[\int_1^3 \left( 3 x^2 + 1 \right)dx\]

\[\int\frac{\sqrt{\tan x}}{\sin x \cos x} dx\]


Solve: (x2 – yx2) dy + (y2 + xy2) dx = 0 


Evaluate `int_(-1)^2 (7x - 5)"d"x` as a limit of sums


If f and g are continuous functions in [0, 1] satisfying f(x) = f(a – x) and g(x) + g(a – x) = a, then `int_0^"a" "f"(x) * "g"(x)"d"x` is equal to ______.


Evaluate the following:

`int_0^pi x sin x cos^2x "d"x`


Evaluate the following:

`int_(pi/3)^(pi/2) sqrt(1 + cosx)/(1 - cos x)^(5/2)  "d"x`


The value of `int_(-pi)^pi sin^3x cos^2x  "d"x` is ______.


The value of `lim_(x -> 0) [(d/(dx) int_0^(x^2) sec^2 xdx),(d/(dx) (x sin x))]` is equal to


Left `f(x) = {{:(1",", "if x is rational number"),(0",", "if x is irrational number"):}`. The value `fof (sqrt(3))` is


The limit of the function defined by `f(x) = {{:(|x|/x",", if x ≠ 0),(0",", "otherwisw"):}`


`lim_(n rightarrow ∞)1/2^n [1/sqrt(1 - 1/2^n) + 1/sqrt(1 - 2/2^n) + 1/sqrt(1 - 3/2^n) + ...... + 1/sqrt(1 - (2^n - 1)/2^n)]` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×