English

Evaluate the following: dee∫02dxex+e-x - Mathematics

Advertisements
Advertisements

Question

Evaluate the following:

`int_0^2 ("d"x)/("e"^x + "e"^-x)`

Sum
Advertisements

Solution

Let I = `int_0^2 ("d"x)/("e"^x + "e"^-x)`

= `int_0^1 ("d"x)/("e"^x + 1/"e"^x)`

= `int_0^1 ("d"x)/(("e"^(2x) + 1)/"e"^x)`

= `int_0^1 ("e"^x "d"x)/("e"^(2x) + 1)`

Put ex = t

⇒ ex dx = dt

Changing the limit, we have

When x = 0

∴ t = e0 = 1

When x = 1

∴ I = `int_1^"e" "dt"/("t"^2 + 1)`

= `[tan^-1 "t"]_1^"e"`

= `[tan^-1 "e" - tan^-1 (1)]`

= `tan^-1 "e" - pi/4`

Hence, I = `tan^-1 "e" - pi/4`.

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 29 | Page 165

RELATED QUESTIONS

Evaluate `int_(-1)^2(e^3x+7x-5)dx` as a limit of sums


Evaluate the following definite integrals as limit of sums. 

`int_2^3 x^2 dx`


Evaluate the definite integral:

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


Evaluate the definite integral:

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


Evaluate the definite integral:

`int_0^1 dx/(sqrt(1+x) - sqrtx)`


Evaluate the definite integral:

`int_0^(pi/2) sin 2x tan^(-1) (sinx) dx`


Evaluate the definite integral:

`int_1^4 [|x - 1|+ |x - 2| + |x -3|]dx`


Prove the following:

`int_0^1 xe^x dx = 1`


Prove the following:

`int_(-1)^1 x^17 cos^4 xdx = 0`


Prove the following:

`int_0^1sin^(-1) xdx = pi/2 - 1`


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


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`


if `int_0^k 1/(2+ 8x^2) dx = pi/16` then the value of k is ________.

(A) `1/2`

(B) `1/3`

(C) `1/4`

(D) `1/5`


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


\[\int\frac{4x + 3}{\sqrt{2 x^2 + 3x + 1}} dx\]

\[\int\sec x \cdot \text{log} \left( \sec x + \tan x \right) dx\]

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

\[\int \sec^4    \text{ x   tan x dx} \]

\[\int\frac{1}{x\sqrt{x^4 - 1}} dx\]

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

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

 


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

Evaluate the following integrals as limit of sums:

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

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


Using L’Hospital Rule, evaluate: `lim_(x->0)  (8^x - 4^x)/(4x
)`


Evaluate `int_1^4 ( 1+ x +e^(2x)) dx` as limit of sums.


Evaluate the following:

`int_0^(pi/2) (tan x)/(1 + "m"^2 tan^2x) "d"x`


Evaluate the following:

`int_0^(1/2) ("d"x)/((1 + x^2)sqrt(1 - x^2))`  (Hint: Let x = sin θ)


If f" = C, C ≠ 0, where C is a constant, then the value of `lim_(x -> 0) (f(x) - 2f (2x) + 3f (3x))/x^2` is


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


`lim_(x -> 0) (xroot(3)(z^2 - (z - x)^2))/(root(3)(8xz - 4x^2) + root(3)(8xz))^4` is equal to


The value of  `lim_(n→∞)1/n sum_(r = 0)^(2n-1) n^2/(n^2 + 4r^2)` is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×