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junior college 1 | H2 Maths
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hi i need help with ii)
Date Posted: 3 years ago
Views: 570
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Pencil is just thought process + me dieing and just writing what I'm thinking.
I wrote 2 methods but I don't recc the second one as it's overly tedious and hence easier to make mistakes. The first method requires you to identify that by transforming the graph by 0.5 units in the positive y-direction u'll get an easier graph to solve. Note that Q wants exact form (albeit annoyingly)
Sorry that I got lazy halfway. Any Qs feel free to ask!
I wrote 2 methods but I don't recc the second one as it's overly tedious and hence easier to make mistakes. The first method requires you to identify that by transforming the graph by 0.5 units in the positive y-direction u'll get an easier graph to solve. Note that Q wants exact form (albeit annoyingly)
Sorry that I got lazy halfway. Any Qs feel free to ask!
Date Posted:
3 years ago
You've forgotten to modulus the integral from x = -2 to x = 0.
-1 + 3e-² is negative
-1 + 3e-² is negative
Ah yea thanks! Made a mistake there
Method without translation of graph :
Integrate y = xeˣ - 0.5 directly from x = -2 to x = 0.5
What we realise is that the portion from x = -2 to the x-intercept is negative (-A - unshaded rectangle) while the portion from the x-intercept to x = 0.5 is positive.
If we put a negative sign outside the integral, we get the opposite. This would be equal to : (A + unshaded rectangle - the other small portion under the curve)
In order to get A + B, we just need to subtract the unshaded rectangle and add in the rectangle from x = 0 to x = 0.5
When x = 0.5 y = 0.5e⁰.⁵ - 0.5
This works out to be :
- ∫⁰.⁵₋₂ (xeˣ - 0.5) dx - (0.5)(2) + (0.5)(0.5e⁰.⁵ - 0.5)
= - [(x-1)eˣ - 0.5x]⁰.⁵₋₂ - 1 + 0.25e⁰.⁵ - 0.25
= - [(0.5 - 1)e⁰.⁵ - 0.5(0.5)] + [(-2 - 1)e-² - 0.5(-2)] - 1.25 + 0.25e⁰.⁵
= 0.5e⁰.⁵ + 0.25 - 3e-² + 1 - 1.25 + 0.25e⁰.⁵
= 0.75e⁰.⁵ - 3e-²
= ¾√e - 3/e²
Integrate y = xeˣ - 0.5 directly from x = -2 to x = 0.5
What we realise is that the portion from x = -2 to the x-intercept is negative (-A - unshaded rectangle) while the portion from the x-intercept to x = 0.5 is positive.
If we put a negative sign outside the integral, we get the opposite. This would be equal to : (A + unshaded rectangle - the other small portion under the curve)
In order to get A + B, we just need to subtract the unshaded rectangle and add in the rectangle from x = 0 to x = 0.5
When x = 0.5 y = 0.5e⁰.⁵ - 0.5
This works out to be :
- ∫⁰.⁵₋₂ (xeˣ - 0.5) dx - (0.5)(2) + (0.5)(0.5e⁰.⁵ - 0.5)
= - [(x-1)eˣ - 0.5x]⁰.⁵₋₂ - 1 + 0.25e⁰.⁵ - 0.25
= - [(0.5 - 1)e⁰.⁵ - 0.5(0.5)] + [(-2 - 1)e-² - 0.5(-2)] - 1.25 + 0.25e⁰.⁵
= 0.5e⁰.⁵ + 0.25 - 3e-² + 1 - 1.25 + 0.25e⁰.⁵
= 0.75e⁰.⁵ - 3e-²
= ¾√e - 3/e²
The translation way would be similar. Integrate directly from x = -2 to x = 0.5, put the negative outside, and add the small rectangle from x = 0 to x = 0.5
When x = 0.5 , y = 0.5e⁰.⁵
Required area (A + B)
= -∫⁰.⁵₋₂ xeˣ dx + (0.5)(0.5e⁰.⁵)
= - [(x-1)eˣ]⁰.⁵₋₂ + 0.25e⁰.⁵
= -[(0.5 - 1)e⁰.⁵] + [(-2 - 1)e-²] + 0.25e⁰.⁵
= 0.5e⁰.⁵ - 3e-² + 0.25e⁰.⁵
= 0.75e⁰.⁵ - 3e-²
= ¾√e - 3/e²
When x = 0.5 , y = 0.5e⁰.⁵
Required area (A + B)
= -∫⁰.⁵₋₂ xeˣ dx + (0.5)(0.5e⁰.⁵)
= - [(x-1)eˣ]⁰.⁵₋₂ + 0.25e⁰.⁵
= -[(0.5 - 1)e⁰.⁵] + [(-2 - 1)e-²] + 0.25e⁰.⁵
= 0.5e⁰.⁵ - 3e-² + 0.25e⁰.⁵
= 0.75e⁰.⁵ - 3e-²
= ¾√e - 3/e²
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