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honeybear101 · Dec 2, 2010, 10:53 PM

I don't know if I got this correct or not. Find lim e^2x/x. x --> +infinity. The answer I got for this was positive infinity. Was that correct?
I also have 2 more question that I can't seem to solve.
(1) What is the area of the largest rectangle under the graph y = 3-x^2 and above the x-axis?
(2) The total area of the region between the curve y=x^3 and the x-axis between x=-1 and x= 2 is?

Unknown008 · Dec 3, 2010, 12:39 AM

Yes, you're right for the first one :)

1.

Let the height of the rectangle be y and the width be x.

The area of the rectangle is then A = xy

But you know that y = 3 - x^2, so you get:

A = x(3 - x^2)

This is the area of the rectangle in terms of x. Expand and simplify. You can find the derivative of this expression to get the value of x at the maximum area, or complete the square and find the optimum point.

2. This is simply:

$\int\ ^2_0\ y\ dx - \int\ ^0_{-1}\ y\ dx$

I put minus because you know the graph is negative for x < 0, and finding the area from -1 to 0 will give a negative area.

Post what you get! :)

honeybear101 · Dec 3, 2010, 01:11 AM

For number 2, I got 9 as my answer. As for number 1, I'm not sure if this is correct, but I got 4 as my answer. >_<

Unknown008 · Dec 3, 2010, 01:26 AM

1. Wait, I realised I made a mistake in my explanation. The width is 2x, not x (though that doesn't really change the answer). This becomes:

$A = 6x - 2x^3$

$A' = 6 - 6x^2$

At optimum, A' = 0.

$0 = 6 - 6x^2$

$x^2 = 1$

$x = 1$

$y = 3 - (1)^2 = 2$

A = 2(1)(2) = 4

Right!

2.

$\int\ ^2_0 x^3\ dx - \int\ ^0_{-1}\ x^3 \ dx = \[\frac{x^4}{4}\]^2_0 - \[\frac{x^4}{4}\]^0_{-1}$

$= \[\frac{(2)^4}{4} - \frac{(0)^4}{4} \]^2_0 - \[\frac{(0)^4}{4} - \frac{(-1)^4}{4}\]^0_{-1}$

honeybear101 · Dec 3, 2010, 08:09 AM

Oh! That's how you do it. Thank you so much for helping me! :)

Unknown008 · Dec 3, 2010, 08:16 AM

No problem! :)