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Applications Of Integration – Beware Of Shady Areas

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Tuition given in the topic of A-Maths Tuition Questions from the tuition centre desk of Miss Loi at 10:55 pm (Singapore time)

Mom and Dad will forever nag about the importance of being a goody two shoes and staying away from shady characters in shady areas. However, the A-Maths Syllabus dictates that every student’s integral-ity be tested by making him/her venture at least once into such areas during the exams. Hur hur hur.

The following question is designed to sieve out the street-smart students from the horribly naive, pampered, and unworldly ones (aren’t we all):

Diagram of Shaded Area Bounded By Curves The diagram shows part of the curve y = 1 + ex, intersecting the y-axis at Q. The tangent to the curve at the point P(1, 1+e) intersects the y-axis at R.

Find the area of the shaded region PQR.

Upon first glance, many of Miss Loi’s students will go: “Walau! So simple! Just integrate the curve to get the area beneath bounded by the points Q and P, and then integrate the straight line to get the area beneath bounded by the same x-coordinates for points R and P, and then subtract the two areas can already!”.

Miss Loi: “So how are you going to integrate that straight line without knowing its equation?”

Student: *frantically flips textbook for chapters on Differentiation and Coordinate Geometry*

頑張って!!!

Revision Exercise

To show that you have understood what Miss Loi just taught you from the centre, you must:

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Comments & Reactions

14 Comments

  1. winston's Avatar
    winston commented in tuition class


    2007
    Jul
    20
    Fri
    6:44am
     
    1

    Wah your first para is sooooooooo lame...Lol!

  2. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    2007
    Jul
    20
    Fri
    10:05am
     
    2

    Winston, you login at 6 in the morning (and risk being late for school) just to say Miss Loi is lame! tsk tsk.

  3. Johnny Malkavian's Avatar
    Johnny Malkavian commented in tuition class


    2007
    Jul
    20
    Fri
    8:16pm
     
    3

    I know how students can sometimes be little monsters, but to describe them as unworldly... hmm.

  4. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    2007
    Jul
    22
    Sun
    10:12am
     
    4

    Depends on your definition of the word.

    From the way we are brought up to the way certain information is 'granted' exposure to the general populace here, can't deny that many of us are 'unworldly', in a certain context of course.

  5. Johnny Malkavian's Avatar
    Johnny Malkavian commented in tuition class


    2007
    Jul
    22
    Sun
    10:19am
     
    5

    It just occured to me, you're the iBook user, aren't you ?

  6. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    2007
    Jul
    23
    Mon
    10:40am
     
    6

    Though something Mac-like has made a guest appearance somewhere in this blog, but ... huh?

  7. uncle sha's Avatar
    uncle sha commented in tuition class


    2007
    Jul
    25
    Wed
    2:40pm
     
    7

    looks like an eqn from my module! ahh i hate that module

    not all students are unworldly lar ... some are sweet and cute like me, keke

  8. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    2007
    Jul
    25
    Wed
    6:29pm
     
    8

    Welcome back encik! Apa kabar?

    You are sweet and cute in an 'unworldly' way *lol*

  9. Kiroii's Avatar
    Kiroii commented in tuition class


    2007
    Jul
    26
    Thu
    8:51pm
     
    9

    dy/dx = m
    m = ex

    You need to sub in x = 1 before you find the equation of the tangent (straight line) at P. Hence the gradient at P, m = e

    equation of line y -(1 + e) = e^x (x - 1)

    Hence straight line equation is:

    y = ex + 1

    at R, x = 0
    with the curve y =e

    intergrate curve with x cords with point 1 and 0
    and we'll get 2.718
    2.718 -( e x 1) - (1/2 x(1 e - 1) x1)
    = 2.718 -e - 1/2(e)
    = 2.718 - 3/2e

    er tats de most i can simply up to-.- whats de ans anyway

    Shaded area = int{0}{1}{(1+e^x)}dx - int{0}{1}{ex+1}dx (or you can use trapezium area formula as you've done)
    = [x + ex]10 - [ (ex2/2) + x]10
    = (e/2 - 1) unit2 (after simplification)
  10. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    2007
    Jul
    27
    Fri
    1:58am
     
    10

    Hello again Kiroii,

    Please see Miss Loi's corrections on your workings! Thanks for your great efforts once more! 😉

  11. Gavin's Avatar
    Gavin  commented in tuition class


    2012
    Mar
    28
    Wed
    7:05pm
     
    11

    The calculations are simpler and more elegant if you shift the whole thing down one unit (which doesn't change the area required). Then the tangent touches the curve at (1,e), and since the gradient is trivially e, the tangent will pass through the origin.

    The required area is therefore int{0}{1}{e^x dx} minus the area of a simple triangle caused by the tangent.

    Easier to see with a graph, obviously.



    • 2012
      Mar
      30
      Fri
      10:19pm
       
      11.1

      @Gavin: Welcome to Jφss Sticks Gavin!

      Must say, your approach certainly radiates elegance and is rather brilliant! The gradient of the line being e, together with P(1, e), would straightaway point to the fact that the newly-shifted line passes through the origin - something Miss Loi missed way back in 2007.

      Oh why didn't she think of this? WHY DIDN'T SHE THINK OF THIS??!!! WHYYYYY????!!!! *stops her hand from slapping herself*

      Alright, the least she could do is present your solution in a proper diagram for the benefit of all:

      Area under the curves alternate solution

      Thanks for reviving and bringing this old blog post from those carefree days in 2007 back to life 😉

  12. Johnny Malkacian's Avatar
    Johnny Malkacian commented in tuition class


    2012
    Mar
    30
    Fri
    10:22pm

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