Advertisements
Advertisements
प्रश्न
Evaluate the following:
`int_0^1 (x"d"x)/sqrt(1 + x^2)`
उत्तर
Let I = `int_0^1 (x"d"x)/sqrt(1 + x^2)`
Put 1 + x2 = t
⇒ 2x dx = dt
⇒ x dx = `"dt"/2`
Changing the limits, we have
When x = 0
∴ t = 1
When x = 1
∴ t = 2
∴ I = `1/2 int_1^2 "dt"/sqrt("t")`
= `1/2 * 2["t"^(1/2)]_1^2`
= `sqrt(2) - 1`
Hence, I = `sqrt(2) - 1`.
APPEARS IN
संबंधित प्रश्न
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_(pi/2)^pi e^x ((1-sinx)/(1-cos x)) dx`
Evaluate the definite integral:
`int_0^(pi/2) (cos^2 x dx)/(cos^2 x + 4 sin^2 x)`
Evaluate the definite integral:
`int_1^4 [|x - 1|+ |x - 2| + |x -3|]dx`
Prove the following:
`int_(-1)^1 x^17 cos^4 xdx = 0`
If f (a + b - x) = f (x), then `int_a^b x f(x )dx` is equal to ______.
Evaluate `int_1^4 ( 1+ x +e^(2x)) dx` as limit of sums.
Evaluate:
`int (sin"x"+cos"x")/(sqrt(9+16sin2"x")) "dx"`
Evaluate `int_(-1)^2 (7x - 5)"d"x` as a limit of sums
If f and g are continuous functions in [0, 1] satisfying f(x) = f(a – x) and g(x) + g(a – x) = a, then `int_0^"a" "f"(x) * "g"(x)"d"x` is equal to ______.
Evaluate the following as limit of sum:
`int _0^2 (x^2 + 3) "d"x`
Evaluate the following as limit of sum:
`int_0^2 "e"^x "d"x`
Evaluate the following:
`int_0^(pi/2) (tan x)/(1 + "m"^2 tan^2x) "d"x`
Evaluate the following:
`int_0^pi x sin x cos^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
The value of `lim_(n→∞)1/n sum_(r = 0)^(2n-1) n^2/(n^2 + 4r^2)` is ______.
`lim_(n→∞){(1 + 1/n^2)^(2/n^2)(1 + 2^2/n^2)^(4/n^2)(1 + 3^2/n^2)^(6/n^2) ...(1 + n^2/n^2)^((2n)/n^2)}` is equal to ______.
