English

Evaluate fd∫-12f(x) dx, where f(x) = |x + 1| + |x| + |x – 1| - Mathematics

Advertisements
Advertisements

Question

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

Sum
Advertisements

Solution

We can redefine f as f(x) = `{{:(2 - x",",  "if" - 1 < x ≤ 0),(x + 2",",  "if"  0 < ≤ 1),(3x",",  "if"  1 < x ≤ 2):}`

Therefore, `int_(-1)^2 "f"(x)"d"x = int_(-1)^0 (2 - x)"d"x + int_0^1 (x + 2)"d"x + int_1^2 3x"d"x`   ....(By P2)

= `(2x = x^2/2)_(-1)^0 + (x^2/2 + 2x)_0^1 + ((3x^2)/2)_1^2`

= `0 - (-2 - 1/2) + (1/2 + 2) + 3(4/2 - 1/2)`

= `5/2 + 5/2 + 9/2`

= `19/2`.

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Integrals - Solved Examples [Page 158]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 12
Chapter 7 Integrals
Solved Examples | Q 19 | Page 158

RELATED QUESTIONS

If `int_0^alpha3x^2dx=8` then the value of α is :

(a) 0

(b) -2

(c) 2 

(d) ±2


If `int_0^alpha(3x^2+2x+1)dx=14` then `alpha=`

(A) 1

(B) 2

(C) –1

(D) –2


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

`int_0^(pi/2)  sqrt(sinx)/(sqrt(sinx) + sqrt(cos x)) 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_0^4 |x - 1| dx`


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


`∫_4^9 1/sqrtxdx=`_____

(A) 1

(B) –2

(C) 2

(D) –1


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


\[\int_\pi^\frac{3\pi}{2} \sqrt{1 - \cos2x}dx\]

Evaluate`int (1)/(x(3+log x))dx` 


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


The total revenue R = 720 - 3x2 where x is number of items sold. Find x for which total  revenue R is increasing.


Prove that `int_0^"a" "f" ("x") "dx" = int_0^"a" "f" ("a" - "x") "d x",` hence evaluate `int_0^pi ("x" sin "x")/(1 + cos^2 "x") "dx"`


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


`int_1^2 1/(2x + 3)  dx` = ______


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


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


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


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


`int_0^(pi/2) 1/(1 + cosx) "d"x` = ______.


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


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


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


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


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


If `intxf(x)dx = (f(x))/2` then f(x) = ex.


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.


If `lim_("n"→∞)(int_(1/("n"+1))^(1/"n") tan^-1("n"x)"d"x)/(int_(1/("n"+1))^(1/"n") sin^-1("n"x)"d"x) = "p"/"q"`, (where p and q are coprime), then (p + q) is ______.


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


`int_0^(2a)f(x)/(f(x)+f(2a-x))  dx` = ______


Evaluate `int_1^2(x+3)/(x(x+2))  dx`


Evaluate: `int_-1^1 x^17.cos^4x  dx`


Evaluate the following integral:

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


Evaluate the following integrals:

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×