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    The diagram shows sketches of the graphs y=2-e^-x and y=x. These graphs intersect at x=a where a>0.
    Use integration to show that the area (between 0 and a and the two lines) is equal to 1+ a - 1/2a^2.

    I can integrate and get 2x + e^-x - 1/2x^2 but I can't get the answer
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    (Original post by Jasminea)
    The diagram shows sketches of the graphs y=2-e^-x and y=x. These graphs intersect at x=a where a>0.
    Use integration to show that the area (between 0 and a and the two lines) is equal to 1+ a - 1/2a^2.

    I can integrate and get 2x + e^-x - 1/2x^2 but I can't get the answer
    So now plug in a into the expression you have, you'll get an answer out in terms of a.

    Now plug in 0 into the expression you have. Subtract this from your previous answer.
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    I tried that but it doesn't work. I got 2a +e^-a - 1/2a^2 - 1
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    do you know how else to do it?
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    (Original post by Jasminea)
    I tried that but it doesn't work. I got 2a +e^-a - 1/2a^2 - 1
    Which is the correct answer for the area between the two graphs. Post a picture of the question, please.
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    But the answer is
    1+ a - 1/2a^2.
    as said above
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    Question 12
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    (Original post by Jasminea)
    Question 12
    Sure. You have 2a +e^{-a} - \frac{a^2}{2} - 1, but you also know that a = 2 - e^{-a}, so: e^{-a} = 2-a, substitute this into your expression for the area.
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    Thanks a lot! My friend and I were getting quite stuck.
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    (Original post by Jasminea)
    Thanks a lot! My friend and I were getting quite stuck.
    If you click on the quotation marks below a post, it lets you quote people so that they know you've replied, just like how you know he's replied. Otherwise they have no clue if you've replied or not (which is why you had to post twice at one stage).

 
 
 
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