Intergrals problem

[Aibetsu]

Hello!
I am having diffuculty starting with this problem:

Use an area to evaluate the intergral from A to A² of 4x(dx) where A > 1.

How do I create a graph of this if the intregral only includes variables? I wanted to see what shape it was so I could solve the problem, but I have no idea how to graph it on my calculator.
Thanks!

 

[galactus]

It would appear you are expected to choose an A. Say, A=2, then A^2=4.

Try those and evaluate. I reckon.

 

[skeeter]

\(\displaystyle \L \int_A^{A^2} 4x dx\) ... \(\displaystyle \L A > 1\)

the area formed by the integral is a trapezoid ...

short base = f(A) = 4A

long base = f(A²) = 4A²

height (along the x-axis) = A² - A

area = (1/2)(A² - A)(4A + 4A²)

area = 2(A² - A)(A² + A) = 2(A⁴ - A²)

graph y = 4x on your calculator and integrate for some A > 1 ... if you have a TI, use the integrate feature on the calculate menu so it will shade in the trapezoid.