help with area problem

[Dennis9]

hi guys.
i need help solving a problem. actually i think the answer is e^7+e^4 but i am not sure.

if anyone has some time can you do it and see if my answer is correct.

find the area between y = (x - 4)e^x and the y-axis from x = 4 to x = 7.

thanks.
dennis

 

[daon]

hi guys.
i need help solving a problem. actually i think the answer is e^7+e^4 but i am not sure.

if anyone has some time can you do it and see if my answer is correct.

find the area between y = (x - 4)e^x and the y-axis from x = 4 to x = 7.

thanks.
dennis

y = (x - 4)e^x
dy = e^x(x-3)dx

Find y limits....
When x = 4, y = 0
When x = 7, y = 3e^7

The area between this curve and the y axis will be the addition of the height of every rectangle (dy) times the length of every rectangle (x) at the limits of y1 (0) to y2 (3e^7)...

A = \(\displaystyle \L \int ^{3e^7} _0 (x) (dy)\) = \(\displaystyle \L \int ^{3e^7} _0 xe^x(x-3)dx\) = \(\displaystyle \L \int ^{3e^7} _0 (x^2e^x-3xe^x)dx\)

 

[soroban]

Hello, Dennis9!

I got a slightly different answer . . .

Find the area between \(\displaystyle y\:=\x - 4)e^x\) and the y-axis from \(\displaystyle x = 4\) to \(\displaystyle x = 7\)

From \(\displaystyle x=4\) to \(\displaystyle x=7\), the graph is on or above the x-axis,
\(\displaystyle \;\;\) so we don't have to worry about "negative area".

We will integrate: \(\displaystyle \L\int^{\;\;\;7}_4 (x-4)e^x\,dx\;\) by parts.

Let: \(\displaystyle \L\,u\,=\,x-4\;\;\;\;dv\,=\,e^x\,dx\)

Then: \(\displaystyle \L\,du\,=\,dx\;\;\;v = e^x\)

We have: \(\displaystyle \L\,(x\,-\,4)e^x\.-\.\int e^x\,dx\;=\;(x\,-\,4)e^x\,-\,e^x\;=\;(x\,-\,5)e^x\)

Evaluate: \(\displaystyle \L\,(x\,-\,5)e^x\bigg]^7_4 \;= \;2e^7\,-\,(-1)e^4 \;= \;2e^7\,+\,e^4\)

 

[daon]

Hello, Dennis9!

I got a slightly different answer . . .

Find the area between \(\displaystyle y\:=\x - 4)e^x\) and the y-axis from \(\displaystyle x = 4\) to \(\displaystyle x = 7\)

From \(\displaystyle x=4\) to \(\displaystyle x=7\), the graph is on or above the x-axis,
\(\displaystyle \;\;\) so we don't have to worry about "negative area".

Soroban, you found the area below the curve, i.e. between (x-4)e^x and the x axis. That was not the question.

 

[Dennis9]

Thanks Soroban. That is what I ended up getting last night. And when I turned it in I got that answer right. Thanks again!

 

[daon]

Thanks Soroban. That is what I ended up getting last night. And when I turned it in I got that answer right. Thanks again!

That doesn't make sense, or at least I am terribly confused. You must have misposted your question, because you found the area between the curve and the x axis.

if y=f(x) > 0 on the given interval then
\(\displaystyle \int^{x2}_{x1} f(x) dx\) is area between f(x) and the x axis.