Maths Integrals
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Let f: (0, 2) → R be defined as f(x) = log?(1 + tan(πx/4)). Then, lim (n→∞) (2/n) [f(1/n) + f(2/n) + . + f(1)] is equal to.
Let f: (0, 2) → R be defined as f(x) = log?(1 + tan(πx/4)). Then, lim (n→∞) (2/n) [f(1/n) + f(2/n) + . + f(1)] is equal to.
A = lim (n→∞) (2/n) ∑ (r=1 to n) f (r/n + n/ (n²)
(The term n/n² seems intended to be part of the function argument, not simply added. The solution proceeds as if it's f (r/n)
A = lim (n→∞) (2/n) ∑ (r=1 to n) [ f (r/n) + f (1/n) + . + f (n-1)/n) ]
The expression in the image seems to be: A = lim (n→∞)
Upvote 
Let f: R → R be a continuous function such that f(x) + f(x + 1) = 2, for all x ∈ R. If I? = ∫(from 0 to 8) f(x)dx and I? = ∫(from -1 to 3) f(x)dx, then the value of I? + 2I? is equal to.
Let f: R → R be a continuous function such that f(x) + f(x + 1) = 2, for all x ∈ R. If I? = ∫(from 0 to 8) f(x)dx and I? = ∫(from -1 to 3) f(x)dx, then the value of I? + 2I? is equal to.
f (x) + f (x + 1) = 2 (1)
replace x with x + 1: f (x + 1) + f (x + 2) = 2 (2)
(2) - (1) ⇒ f (x + 2) = f (x)
∴ f (x) is periodic with period 2.
I? = ∫? f (x)dx = 4 ∫? ² f (x)dx.
I? = ∫? ³ f (x)dx = ∫? f (u-1)du. Let u = x+1.
I? = ∫? f (x-1)dx = 2 ∫? ² f (x-1)dx.
From (1), f (x-1) + f (x) = 2.
I? + 2I? = 4∫?
Upvote If f(a + b + 1 − x) = f(x), for all x, where a and b are fixed positive real numbers, then 1/(a+b) ∫[a,b] x(f(x) + f(x + 1))dx is
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If f(a + b + 1 − x) = f(x), for all x, where a and b are fixed positive real numbers, then 1/(a+b) ∫[a,b] x(f(x) + f(x + 1))dx is
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Answer given by NTS is (1) which is wrong.
I = 1/ (a+b) ∫? x [f (x) + f (x+1)]dx . (1)
Using the property x → a + b - x
I = 1/ (a+b) ∫? (a+b-x) [f (a+b-x) + f (a+b+1-x)]dx
Given f (a+b+1-x) = f (x)
I = 1/ (a+b) ∫? (a+b-x) [f (x+1) + f (x)]dx . (2)
Adding (1) and (2):
2I = 1/ (a+b) ∫? (a+b) [f (x) + f (x+1)
Upvote For real numbers α, β, γ and δ, if ∫ [ (x²-1) + tan⁻¹((x²+1)/x) ] / [ (x⁴+3x²+1)tan⁻¹((x²-1)/x) ] dx = αlogₑ|tan⁻¹((x²+1)/x)| + βtan⁻¹((γ(x²-1))/x) + δtan⁻¹((x²+1)/x) + C where C is an arbitrary constant, then the value of 10(α + βγ + δ) is equal to.
For real numbers α, β, γ and δ, if ∫ [ (x²-1) + tan⁻¹((x²+1)/x) ] / [ (x⁴+3x²+1)tan⁻¹((x²-1)/x) ] dx = αlogₑ|tan⁻¹((x²+1)/x)| + βtan⁻¹((γ(x²-1))/x) + δtan⁻¹((x²+1)/x) + C where C is an arbitrary constant, then the value of 10(α + βγ + δ) is equal to.
The integral is I = ∫ [ (x²-1) + tan? ¹ (x + 1/x)] / [ (x? +3x²+1)tan? ¹ (x+1/x)] dx
This is a complex integral. The provided solution splits it into two parts:
I? = ∫ (x²-1) / [ (x? +3x²+1)tan? ¹ (x+1/x)] dx
I? = ∫ 1 / (x? +3x²+1) dx
The solution proceeds with substitutions which are hard to follow due
Upvote Let P(x) = x² + bx + c be a quadratic polynomial with real coefficients such that ∫₀¹ P(x)dx = 1 and P(x) leaves remainder 5 when it is divided by (x - 2). Then the value of 9(b + c) is equal to:
Let P(x) = x² + bx + c be a quadratic polynomial with real coefficients such that ∫₀¹ P(x)dx = 1 and P(x) leaves remainder 5 when it is divided by (x - 2). Then the value of 9(b + c) is equal to:
P (x) = x² + bx + c.
Given ∫? ¹ P (x) dx = 1.
∫? ¹ (x² + bx + c) dx = [x³/3 + bx²/2 + cx] from 0 to 1 = 1/3 + b/2 + c = 1.
2 + 3b + 6c = 6 => 3b + 6c = 4 - (i)
When P (x) is divided by (x-2), the remainder is 5. So, P (2) = 5.
(2)² + b (2) + c = 5 => 4 + 2b + c = 5 => 2b + c = 1 - (ii)
From (ii), c = 1 -
Upvote 
The problem is to evaluate the integral:
I = ∫? ¹? [x] * e^ [x] / e^ (x-1) dx, where [x] denotes the greatest integer function.
The solution breaks the integral into a sum of integrals over unit intervals:
I = ∑? ∫? ¹ n * e? / e^ (x-1) dx
= ∑? n * e? ∫? ¹ e^ (1-x) dx
= ∑? n * e? [-e^ (1-x)] from n to n+1
Upvote The number of solution of the equation x + 2tanx = π/2 in the interval [0, 2π] is:
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The number of solution of the equation x + 2tanx = π/2 in the interval [0, 2π] is:
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Find the number of solutions for 2tan(x) = π/2 - x in [0, 2π].
This is equivalent to finding the number of intersection points of the graphs y = tan(x) and y = (π/4) - x/2.
Let's sketch the graphs:
y = tan(x) has vertical asymptotes at x = π/2, 3π/2.
y = (π/4) - x/2 is a straight line with a negative sl
Upvote The equation of the planes parallel to the plane x - 2y + 2z - 3 = 0 which are at unit distance from the point (1,2,3) is ax + by + cz + d = 0. If (b-d) = K(c-a), then the positive value of K is ______.
The equation of the planes parallel to the plane x - 2y + 2z - 3 = 0 which are at unit distance from the point (1,2,3) is ax + by + cz + d = 0. If (b-d) = K(c-a), then the positive value of K is ______.
The equation of a plane parallel to x - 2y + 2z - 3 = 0 is x - 2y + 2z + λ = 0.
The distance from the point (1, 2, 3) to this plane is 1.
|1 - 2 (2) + 2 (3) + λ| / √ (1² + (-2)² + 2²) = 1
|1 - 4 + 6 + λ| / √9 = 1
|3 + λ| / 3 = 1
|3 + λ| = 3
3 + λ = 3 or 3 + λ = -3
λ = 0 or λ = -6.
Upvote If the functions are defined as f(x) = √x and g(x) = √(1-x), then what is the common domain of the following functions: f + g, f - g, f/g, g/f, g - f where (f+g)(x) = f(x) + g(x), (f/g)(x) = f(x)/g(x)
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If the functions are defined as f(x) = √x and g(x) = √(1-x), then what is the common domain of the following functions: f + g, f - g, f/g, g/f, g - f where (f+g)(x) = f(x) + g(x), (f/g)(x) = f(x)/g(x)
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f (x) + g (x) = √x + √1-x. The domain requires x ≥ 0 and 1-x ≥ 0, so x ≤ 1. Domain is [0,1].
f (x) - g (x) = √x - √1-x. Domain is [0,1].
f (x)/g (x) = √x / √1-x. Requires x ≥ 0 and 1-x > 0, so x < 1. Domain is [0,1).
g (x)/f (x) = √1-x / √x. Requires 1-x ≥ 0 and x > 0. Domain is (0,1].
The common domain for all these functi
Upvote Which of the following statements is incorrect for the function g(α) for α ∈ R such that g(α) = ∫[from α to (π/2 - α)] (sin x / (cos x + sin x)) dx
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Which of the following statements is incorrect for the function g(α) for α ∈ R such that g(α) = ∫[from α to (π/2 - α)] (sin x / (cos x + sin x)) dx
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Kindly consider the following figure
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