Advertisements
Advertisements
प्रश्न
Evaluate `int_-1^1 |x^4 - x|dx`.
Advertisements
उत्तर
Let I = `int_-1^1 |x^4 - x|dx`
= `int_-1^0 (x^4 - x)dx - int_0^1 (x^4 - x)dx`
= `[x^5/5 - x^2/2]_-1^0 - [x^5/5 - x^2/2]_0^1`
= `[(0 - 0) - ((-1)/5 - 1/2)] - [(1/5 - 1/2) - 0]`
= `7/10 + 3/10`
= 1.
APPEARS IN
संबंधित प्रश्न
By using the properties of the definite integral, evaluate the integral:
`int_0^a sqrtx/(sqrtx + sqrt(a-x)) dx`
Show that `int_0^a f(x)g (x)dx = 2 int_0^a f(x) dx` if f and g are defined as f(x) = f(a-x) and g(x) + g(a-x) = 4.
Evaluate the definite integrals `int_0^pi (x tan x)/(sec x + tan x)dx`
Prove that `int_0^af(x)dx=int_0^af(a-x) dx`
hence evaluate `int_0^(pi/2)sinx/(sinx+cosx) dx`
Evaluate : \[\int(3x - 2) \sqrt{x^2 + x + 1}dx\] .
Evaluate`int (1)/(x(3+log x))dx`
Evaluate : `int "x"^2/("x"^4 + 5"x"^2 + 6) "dx"`
Using properties of definite integrals, evaluate
`int_0^(π/2) sqrt(sin x )/ (sqrtsin x + sqrtcos x)dx`
`int_0^1 (1 - x/(1!) + x^2/(2!) - x^3/(3!) + ... "upto" ∞)` e2x dx = ?
`int_0^(pi"/"4)` log(1 + tanθ) dθ = ______
`int_2^3 x/(x^2 - 1)` dx = ______
`int_0^1 (1 - x)^5`dx = ______.
`int_0^1 x tan^-1x dx` = ______
`int_((-pi)/4)^(pi/4) "dx"/(1 + cos2x)` is equal to ______.
If `f(a + b - x) = f(x)`, then `int_0^b x f(x) dx` is equal to
The value of `int_0^1 tan^-1 ((2x - 1)/(1 + x - x^2)) dx` is
Evaluate: `int_(-1)^3 |x^3 - x|dx`
Evaluate: `int_2^5 sqrt(x)/(sqrt(x) + sqrt(7) - x)dx`
The value of the integral `int_(-1)^1log_e(sqrt(1 - x) + sqrt(1 + x))dx` is equal to ______.
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 ______.
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 ______.
What is `int_0^(π/2)` sin 2x ℓ n (cot x) dx equal to ?
Evaluate: `int_1^3 sqrt(x + 5)/(sqrt(x + 5) + sqrt(9 - x))dx`
For any integer n, the value of `int_-π^π e^(cos^2x) sin^3 (2n + 1)x dx` is ______.
Evaluate the following integral:
`int_0^1 x(1-x)^5 dx`
Evaluate `int_1^2(x+3)/(x(x+2)) dx`
Evaluate the following integral:
`int_0^1 x (1 - x)^5 dx`
The value of \[\int_{-1}^{1}\left(\sqrt{1+x+x^{2}}-\sqrt{1-x+x^{2}}\right)\mathrm{d}x\] is
