Advertisements
Advertisements
Question
By using the properties of the definite integral, evaluate the integral:
`int_2^8 |x - 5| dx`
Solution
`int_2^8 abs (x - 5) dx`
Define,
`abs(x - 5) = {(-(x - 5), if x - 5 < 0, or x< 5),(x - 5, if x - 5 >= 0, or x >=5):}`
`= int_2^5 abs (x - 5) dx + int_2^8 abs (x - 5) dx`
`= - int_2^5 (x - 5) dx + int_2^8 (x - 5) dx`
`= - [x^2/2 - 5x]_2^5 + [x^2/2 - 5x]_5^8`
`= - [25/2 - 25 - 4/2 + 10]`
`= [64/2 - 4 - 25/2 + 25]`
`= - [(-9)/2] + [9/2]`
`= 9/2 + 9/2`
= 9
APPEARS IN
RELATED QUESTIONS
Evaluate : `int e^x[(sqrt(1-x^2)sin^-1x+1)/(sqrt(1-x^2))]dx`
Evaluate : `intlogx/(1+logx)^2dx`
Evaluate `int_(-2)^2x^2/(1+5^x)dx`
Evaluate: `int_(-a)^asqrt((a-x)/(a+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_0^2 xsqrt(2 -x)dx`
Evaluate the definite integrals `int_0^pi (x tan x)/(sec x + tan x)dx`
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that
Evaluate : `int "x"^2/("x"^4 + 5"x"^2 + 6) "dx"`
Evaluate the following integrals : `int_2^5 sqrt(x)/(sqrt(x) + sqrt(7 - x))*dx`
Evaluate the following integral:
`int_0^1 x(1 - x)^5 *dx`
Evaluate `int_0^1 x(1 - x)^5 "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 = ______
`int_0^{pi/2}((3sqrtsecx)/(3sqrtsecx + 3sqrt(cosecx)))dx` = ______
`int_0^{pi/4} (sin2x)/(sin^4x + cos^4x)dx` = ____________
`int_0^pi sin^2x.cos^2x dx` = ______
`int_(-1)^1 log ((2 - x)/(2 + x)) "dx" = ?`
`int_0^{pi/2} (cos2x)/(cosx + sinx)dx` = ______
`int_-1^1x^2/(1+x^2) dx=` ______.
`int_(-pi/4)^(pi/4) 1/(1 - sinx) "d"x` = ______.
`int_0^(2"a") "f"(x) "d"x = 2int_0^"a" "f"(x) "d"x`, if f(2a – x) = ______.
If `int_0^"a" 1/(1 + 4x^2) "d"x = pi/8`, then a = ______.
If `int (log "x")^2/"x" "dx" = (log "x")^"k"/"k" + "c"`, then the value of k is:
Evaluate:
`int_2^8 (sqrt(10 - "x"))/(sqrt"x" + sqrt(10 - "x")) "dx"`
Evaluate: `int_0^(π/2) 1/(1 + (tanx)^(2/3)) dx`
Evaluate: `int_1^3 sqrt(x)/(sqrt(x) + sqrt(4) - x) dx`
Let a be a positive real number such that `int_0^ae^(x-[x])dx` = 10e – 9 where [x] is the greatest integer less than or equal to x. Then, a is equal to ______.
If `β + 2int_0^1x^2e^(-x^2)dx = int_0^1e^(-x^2)dx`, then the value of β is ______.
Let f be continuous periodic function with period 3, such that `int_0^3f(x)dx` = 1. Then the value of `int_-4^8f(2x)dx` is ______.
Evaluate: `int_0^π 1/(5 + 4 cos x)dx`
`int_((-π)/2)^(π/2) log((2 - sinx)/(2 + sinx))` is equal to ______.
`int_0^(π/2)((root(n)(secx))/(root(n)(secx + root(n)("cosec" x))))dx` is equal to ______.
`int_-1^1 (17x^5 - x^4 + 29x^3 - 31x + 1)/(x^2 + 1) dx` is equal to ______.
Evaluate the following limit :
`lim_("x"->3)[sqrt("x"+6)/"x"]`
`int_-9^9 x^3/(4-x^2) dx` =______
Evaluate the following integral:
`int_-9^9 x^3 / (4 - x^2) dx`
Evaluate the following integral:
`int_0^1 x(1 - x)^5 dx`
Evaluate the following integral:
`int_0^1 x (1 - x)^5 dx`
Evaluate the following integral:
`int_0^1x(1 - x)^5dx`
