Advertisements
Advertisements
प्रश्न
By using the properties of the definite integral, evaluate the integral:
`int_2^8 |x - 5| dx`
Advertisements
उत्तर
`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
संबंधित प्रश्न
Prove that: `int_0^(2a)f(x)dx=int_0^af(x)dx+int_0^af(2a-x)dx`
If `int_0^alpha3x^2dx=8` then the value of α is :
(a) 0
(b) -2
(c) 2
(d) ±2
Evaluate :`int_0^pi(xsinx)/(1+sinx)dx`
`int_(-pi/2)^(pi/2) (x^3 + x cos x + tan^5 x + 1) dx ` is ______.
\[\int\limits_0^a 3 x^2 dx = 8,\] find the value of a.
Evaluate : \[\int(3x - 2) \sqrt{x^2 + x + 1}dx\] .
Evaluate : `int "e"^(3"x")/("e"^(3"x") + 1)` dx
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"`
`int_0^(pi/4) (sec^2 x)/((1 + tan x)(2 + tan x))`dx = ?
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_2^3 x/(x^2 - 1)` dx = ______
`int_0^(pi/2) sqrt(cos theta) * sin^2 theta "d" theta` = ______.
`int_0^1 x tan^-1x dx` = ______
`int_3^9 x^3/((12 - x)^3 + x^3)` dx = ______
`int_0^pi sin^2x.cos^2x dx` = ______
`int_0^(pi/2) 1/(1 + cos^3x) "d"x` = ______.
Show that `int_0^(pi/2) (sin^2x)/(sinx + cosx) = 1/sqrt(2) log (sqrt(2) + 1)`
`int_(-2)^2 |x cos pix| "d"x` 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:
`int (dx)/(e^x + e^(-x))` is equal to ______.
Evaluate: `int_0^(π/2) 1/(1 + (tanx)^(2/3)) dx`
Evaluate: `int_0^(2π) (1)/(1 + e^(sin x)`dx
Evaluate: `int_2^5 sqrt(x)/(sqrt(x) + sqrt(7) - 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_4^9 1/sqrt(x)dx` = ______.
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 `int_(-a)^a(|x| + |x - 2|)dx` = 22, (a > 2) and [x] denotes the greatest integer ≤ x, then `int_a^(-a)(x + [x])dx` is equal to ______.
If `β + 2int_0^1x^2e^(-x^2)dx = int_0^1e^(-x^2)dx`, then the value of β is ______.
Evaluate: `int_0^(π/4) log(1 + tanx)dx`.
Evaluate the following integral:
`int_0^1 x(1 - 5)^5`dx
Evaluate:
`int_0^1 |2x + 1|dx`
Solve the following.
`int_0^1 e^(x^2) x^3dx`
Evaluate the following integral:
`int_-9^9x^3/(4-x^2)dx`
Evaluate the following integral:
`int_-9^9 x^3/(4-x^2)dx`
Evaluate the following integral:
`int_0^1x(1-x)^5dx`
Evaluate:
`int_0^sqrt(2)[x^2]dx`
