Area A(x) rectangle under curve

**xxsteelxx** · March 11th 2010, 04:14 PM

Let A(x) be the area of the rectangle inscribed under the curve $y= e^{-2x^{2}}$ with vertices at (-x,0) and (x,0), $x\geq 0$ .

a) Find A(1).
b) What is the greatest value of A(x)? Justify your answer.
c) What is the average value of A(x) on the interval $0 \leq x \leq 2$ ?

I obtained:
a) $A(x)=2x{e}^{-2{x}^{2}}$ ;A(1)= $2{e}^{-2}$
b) $A'(x)=(-4{x}^2 +1)2{e}^{-2{x}^{2}}$ ; Since A increases on $-\frac{1}{2} < x < \frac{1}{2}$ and decreases on $\frac{1}{2} < x < \infty$ , there is a local maximum at $x=\frac{1}{2}$ . $A(\frac{1}{2}) ={e}^{-\frac{1}{2}}$

c)avg value = $\frac{1}{2} \int_0^2 {A(x)} dx = \frac{1}{4}(1-e^{-8})$

I am uncertain if I have found the correct function for A(x).

**skeeter** · March 11th 2010, 04:18 PM

Originally Posted by xxsteelxx

Let A(x) be the area of the rectangle inscribed under the curve $y= e^{-2x^{2}}$ with vertices at (-x,0) and (x,0), $x\geq 0$ .

a) Find A(1).
b) What is the greatest value of A(x)? Justify your answer.
c) What is the average value of A(x) on the interval $0 \leq x \leq 2$ ?

I obtained:
a) $A(x)=2x{e}^{-2{x}^{2}}$ ;A(1)= $2{e}^{-2}$
b) $A'(x)=(-4{x}^2 +1)2{e}^{-2{x}^{2}}$ ; Since A increases on $-\frac{1}{2} < x < \frac{1}{2}$ and decreases on $\frac{1}{2} < x < \infty$ , there is a local maximum at $x=\frac{1}{2}$ . $A(\frac{1}{2}) ={e}^{-\frac{1}{2}}$

c)avg value = $\frac{1}{2} \int_0^2 {A(x)} dx = \frac{1}{4}(1-e^{-8})$

I am uncertain if I have found the correct function for A(x).

your function for A(x) is correct.

**xxsteelxx** · March 11th 2010, 04:20 PM

Thanks, I do not have the solution to this problem and just wanted to make sure.

Thread: Area A(x) rectangle under curve

LinkBack

Thread Tools

Display

Area A(x) rectangle under curve

Similar Math Help Forum Discussions

area of rectangle

Rectangle intersection with curve(Object)

Area of Rectangle

area of a rectangle

Area of a rectangle under a cosine curve?

Search Tags