J's answer to Anonymous's Junior College 1 H2 Maths Singapore question.
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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
Date Posted:
3 years ago
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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