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).

Question

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 =?

Answer 1

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

Answer 2

To find the area between the line and the curve, we need to determine the points of intersection between the line and the curve, and then integrate the difference between the two functions over the interval.

The given line equation is y = 1, and the curve equation is y = 1/√x.

To find the points of intersection, we set the two equations equal to each other and solve for x:

1 = 1/√x

To simplify the expression, we can square both sides:

1 = 1/x

Multiplying both sides by x, we get:

x = 1

So, the line and the curve intersect at x = 1.

To find the other point of intersection, we substitute y = 1 into the curve equation:

1 = 1/√x

Squaring both sides again:

1 = 1/x

Multiplying both sides by x:

x = 1

So, the line and the curve intersect at x = 1 and x = 1.

Now, we can find the area between the line and the curve by integrating the difference between the two functions over the interval [1, 3]:

Area = ∫ (1/√x - 1) dx from x = 1 to 3

To evaluate this integral, we can use the substitution u = √x:

Area = ∫ (1/u - 1) 2u du from u = 1 to √3

Simplifying the integrand:

Area = ∫ (2u/u - 1) du from u = 1 to √3

Area = ∫ (u - 1) du from u = 1 to √3

Integrating:

Area = [(1/2)u^2 - u] from u = 1 to √3

Plugging in the limits of integration:

Area = [(1/2)(√3)^2 - √3] - [(1/2)(1)^2 - 1]

Simplifying:

Area = [3/2 - √3] - [1/2 - 1]

Area = 3/2 - √3 - 1/2 + 1

Area = 3 - √3

So, the area of the region between the line and the curve at x = 1 and x = 3 is 3 - √3.

Answer 3

No problem!

To find the area between the line and the curve, we need to find the definite integral of the difference between the curve and the line over the given interval.

The equation of the line in this case is y = y (which is the same as y = y/1), and the equation of the curve is y = 1/√x.

The area between the line and the curve can be found by evaluating the definite integral:

Area = ∫[(1/√x) - y] dx from x = 1 to 3

Now, let's substitute y = 1/√x into the integral:

Area = ∫[(1/√x) - (1/√x)] dx from x = 1 to 3

Simplifying the integral:

Area = ∫[0] dx from x = 1 to 3
Area = 0

Therefore, the area between the line and the curve at x=1 and x=3 is 0. It seems there might be a mistake in the original question or calculation. Please double-check your inputs and equations to clarify.