How do you calculate area under the curve of a function?

Question

A graph is given for f(x)= square root of x...shows a triangle under the graph with one leg as x-axis and other at (4,0) to a point on the curve.

An engineer estimates the area under the curve of the function f(x)=square root of x from x= 0 to x = 4 unsing the triangle shown. What is the difference between the estimate and the actual value of the area?

Answer 1

The definite integral of the function from x=0 to x=4 is the area under the curve between those points

The integral of x^(1/2) is [2/3]*x^(3/2) + k

Evaluate this for the definte interval of x=0 to x=4, remembering that k -k = 0' .

The value at x=0 is zero so we only need to evaluate it at x=4

[2/3]*x^(3/2)
[2/3]*4^(3/2)
[2/3]*8
16/3
5.333333....

The difference between the estimate and the actual value is 5.33333.... - 4 = 1.33333...

There you go!

Answer 2

You use higher math called Integral Calculus to get an exact area under a curve.

Here is the formula to use (I did the integration for you)

F(x) = 2/3 * x^3/2

^3/2 means 1.5th root... use the y^x key on your calculator... y is 1.5 ... x is 0 and 4

Take F(x) at 4, then subtact F(x) at 0 ((which is 0) -- that's your answer.

I get 5.333333

The difference between the triangle estimate and this -- you can figure out easy, for yourself.


.

Answer 3

The area under the curve = integration from 0 to 4 f(x) . dx
= [ (2/3) * x^(3/2) ] = 16/3
The engineer estimated area = 4 square units
The difference = 4/3 s.u

Answer 4

What you do is "slice" the sphere into many circular slices of infinitesimal length. Now, each half circle is a curve which can be integrated by the method mentioned to get their area. By integrating again (double integral), over all these infinitesimals areas, you get volume.

Answer 5

you intergrate the equation: (raise the power of the x up by one and the divide by new power) so x^2 = 1/3x^3.
Then put 4 into the new intergration and then 0 and then f(4) - f(0) and you should have the area.

Or you can use the trapezium rule... too complex to explain.

1/2 x h x [(y0 + yn) + 2 x (y1 + y2 + y3 +.... yn-1)]

Answer 6

You would use integration.

In this case, it would be S[0,4] sqrt (x) dx - **estimate of triangle's area**

That integral is (2/3)(4)^(3/2) - (2/3)(3)^(3/2)

Answer 7

Calculus!

You must integrate the area under the curve...

Answer 8

you find the two end points of the area you want to get and then take the derivative of that area.