Maths Integrals

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2 months ago

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R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image 

 

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V
Vishal Baghel

Contributor-Level 10

∫? ² |x-1|-x|dx
Let f (x)=|x-1|-x|
= {|1-2x|, x≤1; 1, x≥1}
A = 1/2+1=3/2
∫? ¹/² (1-2x)dx+∫? /? ¹ (2x-1)+∫? ²1dx
= [x-x²]? ¹/²+ [x]? ² = 3/21dx
= [x-x²]? ¹/²+ [x]? ² = 3/2

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A
alok kumar singh

Contributor-Level 10

|z|-Re (z)≤1 ⇒ √ (x²+y²)-x≤1 ⇒ √ (x²+y²)≤1+x ⇒ y²≤1+2x=2 (x+1/2).

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R
Raj Pandey

Contributor-Level 9

I = ∫ (π/24 to 5π/24) dx/ (1+³√tan2x). Using King's rule.
2I = ∫ (π/24 to 5π/24) dx = 4π/24=π/6. I=π/12.

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R
Raj Pandey

Contributor-Level 9

f (x) = ∫? [y]dy. For x∈ [n, n+1), [y]= [x]=n.
f (x) = Σ (k=0 to n-1) ∫? ¹ k dy + ∫? n dy = Σk + n (x-n).
f (x) is continuous at integers. f' (x)=n= [x]. Not differentiable at integers.

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2 months ago

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A
alok kumar singh

Contributor-Level 10

2(cos²θ/sin²θ) - 5/sinθ + 4 = 0
(2sinθ - 1)(sinθ - 2) = 0
sinθ = 1/2 only
∴ θ = π/6, 5π/6
↓↓
θ?, θ?
∫(π/6 to 5π/6) cos²3θdθ = ∫(π/6 to 5π/6) (1+cos6θ)/2 dθ = π/3

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2 months ago

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R
Raj Pandey

Contributor-Level 9

sinx = t, cosxdx = dt
I = ∫dt/(t³(1+t?)^(2/3)) = ∫dt/(t?(1+1/t?)^(2/3))
Put 1+1/t? = r³ ⇒ -6/t? dt = 3r²dr
I = (-1/2)∫dr = -1/2 r + c = (-1/2)(1+1/sin?x)^(1/3) + c
f(x) = (-1/2)cosec²x(1+sin?x)^(1/3) and λ=3
⇒ f(π/3) = -2

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