Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
|
One Answer Below
Anyone can contribute an answer, even non-tutors.
I couldn't answer this question
Date Posted: 3 years ago
Views: 317
See 1 Answer
You will need to integrate by parts and also use the fact that ∫ sinh x dx = cosh x + constant.
Keep x² intact and integrate sinh x for the first 'by parts' and subsequently, keep the term in x and integrate sinhx again.
∫¹₀ x² sinh x dx
= x² cosh x - ∫¹₀ 2x cosh x dx
= x² cosh x - ( 2x sinh x - ∫¹₀ 2 sinh x dx)
= [x² cosh x - 2x sinh x + 2 cosh x]¹₀
= [(x² + 2) cosh x - 2x sinh x]¹₀
= [(1² + 2) cosh 1 - 2(1) sinh 1] - [(0² + 2) cosh 0 - 2(0) sinh 0]
= 3 cosh 1 - 2 sinh 1 - 2 cosh 0
= 3 (e¹ + e-¹)/2 - 2(e¹ - e-¹)/2 - 2(e⁰ + e-⁰)/2
= 3/2 e + 3/2e - e + 1/e - 2
= ½e + 5/2e - 2
≈ 0.27883952
Keep x² intact and integrate sinh x for the first 'by parts' and subsequently, keep the term in x and integrate sinhx again.
∫¹₀ x² sinh x dx
= x² cosh x - ∫¹₀ 2x cosh x dx
= x² cosh x - ( 2x sinh x - ∫¹₀ 2 sinh x dx)
= [x² cosh x - 2x sinh x + 2 cosh x]¹₀
= [(x² + 2) cosh x - 2x sinh x]¹₀
= [(1² + 2) cosh 1 - 2(1) sinh 1] - [(0² + 2) cosh 0 - 2(0) sinh 0]
= 3 cosh 1 - 2 sinh 1 - 2 cosh 0
= 3 (e¹ + e-¹)/2 - 2(e¹ - e-¹)/2 - 2(e⁰ + e-⁰)/2
= 3/2 e + 3/2e - e + 1/e - 2
= ½e + 5/2e - 2
≈ 0.27883952
360 Tutoring Program