Calc: area of rectangle under curve

[rjs123]

Homework Statement

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 >= 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 <= x <= 2

The Attempt at a Solution

a. Logically i can just plug 1 or -1 into the equation to find its y intercept and then find the area of the rectangle I get A(1) = e^(-2) * 2 or .27 using the first method.

...but I am pretty sure there is a calculus method...here is my attempt...not sure if I am on the right track.

y = e^-2x^2
lny = lne^-2x^2
lny = -2x^2
take derivative
1/y = -4x

b. the graph seems to extend from negative infinity to postive infinity so I'm not seeing a greatest value

c.

 

[Mark44]

Homework Statement

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 >= 0

The vertices are a little confusing. Did you mean (-x, 0) for the first one?

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 <= x <= 2

The Attempt at a Solution

a. Logically i can just plug 1 or -1 into the equation to find its y intercept and then find the area of the rectangle I get A(1) = e^(-2) * 2 or .27 using the first method.

This seems fine. Finding A(1) doesn't require calculus.

...but I am pretty sure there is a calculus method...here is my attempt...not sure if I am on the right track.

y = e^-2x^2
lny = lne^-2x^2
lny = -2x^2
take derivative
1/y = -4x

b. the graph seems to extend from negative infinity to postive infinity so I'm not seeing a greatest value

The problem isn't asking about the greatest value of y = e^-2x2. It's asking about the greatest value of A(x). Find a formula for A(x) and then do the usual thing with finding A'(x) and setting it to zero.

c.

 

[rjs123]

The vertices are a little confusing. Did you mean (-x, 0) for the first one?
This seems fine. Finding A(1) doesn't require calculus.
The problem isn't asking about the greatest value of y = e^-2x2. It's asking about the greatest value of A(x). Find a formula for A(x) and then do the usual thing with finding A'(x) and setting it to zero.

ok so for b I've attempted this but I am stuck

$$y = e^{-2x^2}$$

$$A = x(e^-2x^2)$$

$$A = xe^-2x^2$$

$$A' = x * -4xe^-2x^2 + 1 * e^-2x^2$$

$$A' = -4x^2*e^-2x^2 + e^-2x^2$$

$$0 = -4x^2*e^-2x^2 + e^-2x^2$$

 

[Mark44]

Factor e^-2x2 out of the two expressions.

 

[Mark44]

ok so for b I've attempted this but I am stuck

$$y = e^{-2x^2}$$

$$A = xe^{-2x^2}$$

$$A = xe^{-2x^2}$$

$$A' = x * -4xe^{-2x^2} + 1 * e^{-2x^2}$$

$$A' = -4x^2*e^{-2x^2} + e^{-2x^2}$$

$$0 = -4x^2*e^{-2x^2} + e^{-2x^2}$$

This is a lot easier to read - thanks!

In your LaTeX, if an exponent is more than 1 character, put braces {} around it. I fixed the above to make them look better.

 

[rjs123]

This is a lot easier to read - thanks!

In your LaTeX, if an exponent is more than 1 character, put braces {} around it. I fixed the above to make them look better.

$$0 = e^{-2x^2}(-4x^2 + 1)$$

$$0 = e^{-2x^2}(-2x + 1)(2x + 1)$$

$$x = -1/2, 1/2$$

width = 1
length =
$$y = e^{-2(1/2)^2}$$
$$y = e^{-2(1/4)}$$
$$y = e^{-1/2}$$

length = .606

A = 1 * .606, A = .606

 

[Mark44]

OK, these are values of x for which A'(x) = 0. Do you get a maximum, minimum, or neither at each of them?

 

[rjs123]

OK, these are values of x for which A'(x) = 0. Do you get a maximum, minimum, or neither at each of them?

I'm still confused on the answer for b since the area for a rectangle under the graph could be infinite since the graph never touches the x-axis.

 

[Mark44]

For each value of x, you get a value A(x), which is the area of one rectangle. As x gets larger, the rectangles get wider, but they also get shorter, which also affects their area.

 

[rjs123]

For each value of x, you get a value A(x), which is the area of one rectangle. As x gets larger, the rectangles get wider, but they also get shorter, which also affects their area.

I used some numbers to justify this...and you are right...if i pick anything greater than or less than 1/2 for x...the area gets exponentially smaller.

Almost done...for c.)... what is the average value of A(x) on the interval 0 <= x <= 2...

should i just use average value theorem...basically get the area when x = 2...divide that by 2?

 

[Mark44]

Be careful when you say area, since you're not really dealing with the area under the curve, but instead the average area of the rectangles. Remember that A(x) gives you the area of a rectangle whose width is 2x and whose height is e^-2x2, for x between 0 and 2.

 

[rjs123]

Be careful when you say area, since you're not really dealing with the area under the curve, but instead the average area of the rectangles. Remember that A(x) gives you the area of a rectangle whose width is 2x and whose height is e^-2x2, for x between 0 and 2.

when 0<=x<=2

b = 2 a = 0

A(b) - A(a)/ b - a

.00134 - 0/ 2 = .006709
This is basically the average area of the rectangle when x is between 0 and 2 correct?

 

[rjs123]

Be careful when you say area, since you're not really dealing with the area under the curve, but instead the average area of the rectangles. Remember that A(x) gives you the area of a rectangle whose width is 2x and whose height is e^-2x2, for x between 0 and 2.

any help on setting this up properly...