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secondary 4 | A Maths
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Can anyone solve this question? I only managed to get A (3,0), and B (0, ln2). I’m unsure about my coordinates of C, which I got C (0,-6) and part iv and v.
2y = ln (4 - x)
4 - x = e^2y
x = 4 - e^2y
k = 2
Notice that the shaded area is made up of a right angled triangle and another part bordered by the axis and the curve.
We use the y-coordinate of B as our upper bound , and y-coordinate of A as our lower bound
∫(upper bound ln2, lower bound 0) x dy
= ∫(ln2, 0) (4 - e^2y) dy
= [4y - ½e^2y] (upper bound ln2, lower bound 0)
= (4ln2 - ½e^(2ln2)) - (4(0) - ½e^(2 x 0) )
= 4ln2 - ½e^ln(2²) + ½
= 4ln2 - ½(4) + ½
= 4ln2 - 3/2
9 + 4ln2 - 3/2
= 4ln2 + 7½
= 4ln2 + 7.5
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