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junior college 2 | H2 Maths
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For question 4b), is there a nd to split the indices or just integrate from the question? I’m kinda stuck, nd help!
Date Posted: 3 years ago
Views: 475
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done
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clear
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Like this perhaps? Seems like it need not be split after all
Date Posted:
3 years ago
Thxs alot for the help!
The expression is directly integrable.
Recall that : d/dx aᶠ⁽ˣ⁾ = f'(x) · aᶠ⁽ˣ⁾ · ln a
So, you can just split the indices like this.
∫¹₀ 2ˣ⁺²^ˣ dx
= ∫¹₀ 2ˣ · 2²^ˣ dx
Then, introduce back in a ln2 for each of the terms.
= (1/ln2)² ∫¹₀ 2ˣ ln2 · 2²^ˣ ln2 dx
Whereby your a = 2, f(x) = 2ˣ and f'(x) = 2ˣ ln2
= 1/(ln2)² [2²^ˣ]∫¹₀
= 1/(ln2)² (2²^¹ - 2²^⁰)
= 2/(ln2)²
(Since 2²^¹ - 2²^⁰ = 2² - 2¹ = 4 - 2 = 2)
The question tests your ability to visualise/realise that it is directly integrable and can be rewritten.
Recall that : d/dx aᶠ⁽ˣ⁾ = f'(x) · aᶠ⁽ˣ⁾ · ln a
So, you can just split the indices like this.
∫¹₀ 2ˣ⁺²^ˣ dx
= ∫¹₀ 2ˣ · 2²^ˣ dx
Then, introduce back in a ln2 for each of the terms.
= (1/ln2)² ∫¹₀ 2ˣ ln2 · 2²^ˣ ln2 dx
Whereby your a = 2, f(x) = 2ˣ and f'(x) = 2ˣ ln2
= 1/(ln2)² [2²^ˣ]∫¹₀
= 1/(ln2)² (2²^¹ - 2²^⁰)
= 2/(ln2)²
(Since 2²^¹ - 2²^⁰ = 2² - 2¹ = 4 - 2 = 2)
The question tests your ability to visualise/realise that it is directly integrable and can be rewritten.
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