# Thread: Area A(x) rectangle under curve

1. ## Area A(x) rectangle under curve

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).
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).

2. 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).
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.

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