• Apr 9th 2006, 02:51 AM
ify00
The area under the graph y= 4 + 2x from x=0 to x=4 is divided into two equal parts by the line x=k. Find the value of k.

Please ive been stuck on this for years
• Apr 9th 2006, 04:02 AM
rgep
You're asking for a value of k such that $\displaystyle \int_0^k 4+2x \, dx = \frac12 \int_0^4 4+2x \, dx$. Doing the integration, $\displaystyle \left[4x+x^2\right]_0^k = \frac12 \left[4x+x^2\right]_0^4$, that is, $\displaystyle k^2 + 4k = \frac12 (4.4+4^2) = 16$. This is a quadratic equation in k which has a root near 2.47.
• Apr 9th 2006, 04:02 AM
earboth
Quote:

Originally Posted by ify00
The area under the graph y= 4 + 2x from x=0 to x=4 is divided into two equal parts by the line x=k. Find the value of k.

Please ive been stuck on this for years

Hello,

I assume that 0 < k < 4.

From the title of your post I assume that you want to calculate the area by integration (which isn't necessary with your problem). You calculate 2 areas which should be equal:
$\displaystyle A_1=\int^{k}_{0}(2x+4)dx=\left[x^2+4x \right]^{k}_{0}=k^2+4k-0$

$\displaystyle A_2=\int^{4}_{k}(2x+4)dx=\left[x^2+4x \right]^{4}_{k}=32-(k^2+4k)$

It is now

$\displaystyle A_1=A_2 \Longrightarrow k^2+4k=32-(k^2+4k)$

You'll get a quadratic equation:

$\displaystyle 2k^2+8k-32=0 \Longleftrightarrow k=-2+2\cdot \sqrt{5}\ \vee \ k= -2-2\cdot \sqrt{5}$

The negative solution doesn't fit into your problem. So k = 2.47...

Greetings

EB