What is the Integral of f(x)g'(x) for Given Values of g(x)?

[Spartan029]

Homework Statement

Estimate \int_{0}^{10} f(x) g'(x) dx for f(x) = x^{2}
and g has the values in the following table.

<br /> \begin{array}{l | c|c|c|c|c|c |} <br /> \hline<br /> \hline g&amp;0&amp;2&amp;4&amp;6&amp;8&amp;10\\<br /> \hline g(x)&amp;2.3&amp;3.1&amp;4.1&amp;5.5&amp;5.9&amp;6.1\\<br /> \hline<br /> \end{array}<br />

Homework Equations

\int uv&#039; dx = uv = \int u&#039;v dx

The Attempt at a Solution

Okay so, since f(x) is x squared i chose

u = x^{2} and v' = g'(x)
&
u' = 2x dx and v = g(x)

plugging in...

g(x)x^{2} - \int_{0}^{10} 2xg(x) dx

and this is where I am stuck. I can't plug in the g values because i first need to take the integral of 2xg(x) ...I think. lol

a nudge in the right direction would be ub3r helpful and much appreciated. thanks!

ps. that latex table took me like a half hour to figure out rofl

 

[gabbagabbahey]

Well, the integral

\int_0^{10} 2xg(x)dx

gives the area under the curve 2xg(x) between x=0 and x=10.

You are given g(x)at certain points along the interval, so what is 2xg(x) at those points? Draw a picture and see if you can find a way to estimate the area under 2xg(x)

 

[Spartan029]

\int_0^{10} 2xg(x)dx

when x= 0, 2xg(x) = 0
x=2, 2xg(x) = 2(2)(3.1) = 12.4
x=4, 2xg(x) = ... = 32.8
x=6, 2xg(x) = ... = 66
x=8, 2xg(x) = ... = 94.4
x=10, 2xg(x) = ... = 122

connect thesse and estimate area under from 0 to 10?
makes sense, but is there any other way to solve the problem?

 

[gabbagabbahey]

Because you are only given a few values of g(x) and not an explicit functional no exact solution will be possible.

You might be able to get a slightly more accurate value by fitting a 4th degree polynomial to the points you are given, but it will still just be an approximation and I don't think your instructor is looking for anything that complicated.

 

[Spartan029]

okay awesome! thanks for helping me out!

 

[Spartan029]

oh wait how do i work in that g(x)x^{2} part

 

[gabbagabbahey]

oh wait how do i work in that g(x)x^{2} part

You mean
g(x)x^2|_0^{10}

right?

Remember, integration by parts means that uv is evaluated at the endpoints of your integration interval.

 

[Spartan029]

You mean
g(x)x^2|_0^{10}

right?

Remember, integration by parts means that uv is evaluated at the endpoints of your integration interval.

geez this problem is pwning me lol.
so we go...

g(x)x^2 - \int_0^{10} 2xg(x)dx
(2.3)(0) - 2(0)(2.3) = 0, for x=0
(3.1)(4) - 2(2)(3.1) = 0, for x=2
(4.1)(16) - 2(4)(4.1) = 54.4, for x=4
(5.5)(36) - 2(6)(5.5) = 132, for x=6
...and so on...

calculate area under (connected) points (0,0) (0,0) (4, 54.4) (6, 132) ...?