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    Name:  hiya.GIF
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    honest to god, i think i've tried every flaming possible method to answering this question and each time i've gotten it wrong.

    Please can i have some help on this question?

    Methods i've tried

    ->Seperate integration i tried integrating each part seperatly and then adding the 2 areas together. Didnae work.

    ->Rearranging to get x because they're y coordinates, i tried rearranging the equation so i could sub in the y coordinates. that didn't work either?

    ->Using x i tried integrating with (1/2,0) using x coordinates but that didn't work either.

    Idk I'm really panicing that i can't answer this, it's due tommorow and i have a test, help needed thank you!!


    btw the answer should be (1/12 + 1/2e )
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    (Original post by ZiggyStardust_)

    honest to god, i think i've tried every flaming possible method to answering this question and each time i've gotten it wrong.

    Please can i have some help on this question?

    Methods i've tried

    ->Seperate integration i tried integrating each part seperatly and then adding the 2 areas together. Didnae work.

    ->Rearranging to get x because they're y coordinates, i tried rearranging the equation so i could sub in the y coordinates. that didn't work either?

    ->Using x i tried integrating with (1/2,0) using x coordinates but that didn't work either.

    Idk I'm really panicing that i can't answer this, it's due tommorow and i have a test, help needed thank you!!


    btw the answer should be (1/12 + 1/2e )
    Not sure why your first method didn't work unless you added the incorrect areas together. The correct answer is given by \displaystyle \int_{0}^{1/2} (1-2x)^5 .dx + \int_{0}^{1/2} -(e^{2x-1}-1) .dx

    Why the negative? Because the latter integral is negative as it's below the y-axis, but we're interested in the positive quantity of it since area is defined to be positive. - so the negative sign turns the negative quantity positive, before we add the two.

    The above doesn't yield the question's answer so I disagree with it, but that's the method.

    EDIT: I think you meant it as \frac{1}{12}+\frac{1}{2e} and not \frac{1}{12}+\frac{1}{2}e? In which case, the two answers are equivalent, you just need to rearrange it in the wanted form. Also means you should probably use some brackets if you don't want people misinterpreting what you write...
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    (Original post by RDKGames)
    Not sure why your first method didn't work unless you added the incorrect areas together. The correct answer is given by \displaystyle \int_{0}^{1/2} (1-2x)^5 .dx + \int_{0}^{1/2} -(e^{2x-1}-1) .dx

    Why the negative? Because the latter integral is negative as it's below the y-axis, but we're interested in the positive quantity of it since area is defined to be positive. - so the negative sign turns the negative quantity positive, before we add the two.

    The above doesn't yield the question's answer so I disagree with it, but that's the method.

    EDIT: I think you meant it as \frac{1}{12}+\frac{1}{2e} and not \frac{1}{12}+\frac{1}{2}e? In which case, the two answers are equivalent, you just need to rearrange it in the wanted form. Also means you should probably use some brackets if you don't want people misinterpreting what you write...
    LOL so I forgot to integrate the curves. I was subbing in the coordinates into the original equation (silly). the amount of dumb mistakes i make in maths is laughable....

    this method makes plenty of sense though, i did it slightly differently tho, i used the y coordinates to integrate separately. anyhow, thank you very very much for providing the method, i think i might actually understand now. I appreciate it

    Also yes, i meant the second one,, that's the answer. Alright I'll use brackets better next time. thank you!!! )
 
 
 
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