I was the one who attempted to answer this for you yesterday.
So, what you are saying is that the line cuts y = 1/√x at (1,1) and (3,1/√3)
the slope of that line is (1/√3 - 1)/2 = (√3 - 3)/6 after simplifying and rationalizing.
equation: y-1 = (√3 - 3)/6 (x - 1)
y = (√3 - 3)/6 (x - 1) + 1
So the area = ∫((√3 - 3)/6 (x - 1) + 1 - 1/√x) dx from 1 to 3
The first part is a linear integration and the second part I did yesterday as 2√x, I will leave it up to you to integrate the first part.
Anyway, I entered the above into Wolfram and got your anticipated answer:
https://www.wolframalpha.com/input/?i=%E2%88%AB%28+%28%E2%88%9A3+-+3%29%2F6+%28x+-+1%29+%2B+1+-+1%2F%E2%88%9Ax%29+dx+from+1+to+3
There is a graph of y=1/sqrtx.Show that the area of the region between line and curve at x=1 and x=3 is 3-(5sqrt3/3).
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asked by Raj
yesterday at 6:35am
"Show that the area of the region between line and curve" . What line?
Assuming you want the area between the curve and the x-axis for the given interval ...
Area = ∫ 1/√x dx from x = 1 to 3
= [2√x] from 1 to 3
= 2√3 - 2√1
= 2√3 - 2, which is not the expected answer.
Check your typing
Sorry for fuzzy thing.The line cuts the curve of 1/(x^1/2) at (1,y) and (3,y).At two different points.Area between line and curve =?
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