Integration help

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MathsHelpPlz

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Dec 13, 2012
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"1. There are graphs y=2-(e^(-x)) and y=x, that intersect at x=a where a>0.
a) Find an equation statisfied by a,
b) Write down an integral which is equal to the area of the shaded region,
c) use integration to show that the area is equal to 1 + a - (1/2)a^2"

So I got a) 2-(e^(-a))=a and b) ∫from a to 0 (2 - e^(-x) - x) dx and both were correct but I am stuck on c). Any help would be much appreciated, thank you for your time.
 
MathsHelpPlz said:
"1. There are graphs y=2-(e^(-x)) and y=x, that intersect at x=a where a>0.
a) Find an equation statisfied by a,
b) Write down an integral which is equal to the area of the shaded region,
c) use integration to show that the area is equal to 1 + a - (1/2)a^2"

So I got a) 2-(e^(-a))=a and b) ∫from a to 0 (2 - e^(-x) - x) dx and both were correct but I am stuck on c). Any help would be much appreciated, thank you for your time.
Click to expand...

*The graphs don't go under the x-axis.
 
c) We have:

\(\displaystyle \displaystyle A=\int_0^a 2-e^{-x}-x\,dx=\left[2x+e^{-x}-\frac{x^2}{2} \right]_0^a=\left(2a+e^{-a}-\frac{a^2}{2} \right)-\left(2(0)+e^0+\frac{0^2}{2} \right)=2a+e^{-a}-\frac{a^2}{2}-1\)

Now, from part b) we know that:

\(\displaystyle e^{-a}=2-a\)

So, substitute for \(\displaystyle e^{-a}\) to get the desired result.
 
Last edited:
MathsHelpPlz said:
"1. There are graphs y=2-(e^(-x)) and y=x, that intersect at x=a where a>0.
a) Find an equation statisfied by a,
b) Write down an integral which is equal to the area of the shaded region,
c) use integration to show that the area is equal to 1 + a - (1/2)a^2"

So I got a) 2-(e^(-a))=a and b) ∫from a to 0 (2 - e^(-x) - x) dx and both were correct but I am stuck on c). Any help would be much appreciated, thank you for your time.
Click to expand...

Are you saying that you cannot integrate the following:

\(\displaystyle \int_0^a{2}dx \ - \ \int_0^a{e^{-x}}dx \ - \ \int_0^a{x}dx\)
 
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