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integration and area (1 Viewer)

pasta8

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how do you solve the area bounded by the curve y= ln(x+4) the y-axis, y=0 and y=1 because i got (e-5) units2 but the solution says (5-e)units 2

is this bit right x+4 = e^y from y=ln(x+4)?

be grateful for anyone who can help.
 
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are you sure its from

y=0 ---> y= 1 (cause you get e-5),

but if its y=1 ---> y=0 (then you get 5-e)
 

Drongoski

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y = ln(x+4) ==> x = ey - 4


 
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Drongoski

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why does area = (0-x) ??/
Hard to explain without diagram.

If u know the graph of y = log x which cuts the x-axis @ x=1, the graph of y = log(x+4) is this graph moved 4 units left. The extent y=log(x+4) is left of the y-axis [which is x=0] is 0 - x where x = e^y - 4.
 

pasta8

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thanks heaps i sort of get it.

In an exam situation could i just find the absolute value of it? because the answer i got was negative but is this always the case when you don't make the area (x-0)dy?
 

Drongoski

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thanks heaps i sort of get it.

In an exam situation could i just find the absolute value of it? because the answer i got was negative but is this always the case when you don't make the area (x-0)dy?
for such case, use abs value. but if multiple regions, treat each separately. good luck.
 

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