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.

$$\begin{array}{l | c|c|c|c|c|c |} \hline \hline g&0&2&4&6&8&10\\ \hline g(x)&2.3&3.1&4.1&5.5&5.9&6.1\\ \hline \end{array}$$

Homework Equations

$$\int uv' dx = uv = \int u'v dx$$

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) ...?